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IB Chemistry R2.1: How much? The amount of chemical change worked solutions

Worked solutions to every practice question in R2.1: How much? The amount of chemical change of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

[1 mark]
Copper(I) oxide reacts with carbon to give copper and carbon monoxide. Which sequence of smallest whole-number coefficients balances \mathrm{Cu}_{2}\mathrm{O}+\mathrm{C}\rightarrow\mathrm{Cu}+\mathrm{C}\mathrm{O}?
A 1, 1, 2, 1 B 1, 1, 1, 1 C 2, 1, 2, 1 D 1, 2, 2, 2
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Cu2O+C→2Cu+CO. Two copper atoms must appear on each side; the other counts are already one each.

Final answer

A

Question B2

[3 marks]
Iron oxyhydroxide decomposes on heating according to the supplied equation: 2\,\mathrm{Fe}\mathrm{O}\mathrm{O}\mathrm{H}\mathrm{(s)}\rightarrow\mathrm{Fe}_{2}\mathrm{O}_{3}\mathrm{(s)}+\mathrm{H}_{2}\mathrm{O}\mathrm{(g)}
(a) [ 2 ] {Calculate the amount of iron(III) oxide formed when 0.0360\,\text{mol} of iron oxyhydroxide decomposes completely.}
(b) [ 1 ] {State the physical state of the water as represented in this equation.}
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The product coefficient is half the FeOOH coefficient: 0.0360/2=0.0180 mol [2: ratio and result]. The supplied (g) identifies gaseous water [1].

Final answer

(a) 0.0180 mol Fe2O3; (b) water vapour

Question B3

[1 mark]
For the supplied reaction below, the initial amounts are 0.120\,\text{mol} ammonia and 0.140\,\text{mol} oxygen. 4\,\mathrm{N}\mathrm{H}_{3}\mathrm{(g)}+5\,\mathrm{O}_{2}\mathrm{(g)}\rightarrow4\,\mathrm{N}\mathrm{O}\mathrm{(g)}+6\,\mathrm{H}_{2}\mathrm{O}\mathrm{(g)}
Which statement correctly identifies the limiting reactant?
A Ammonia, because 0.120/5 is less than 0.140/4. B Ammonia, because 0.120/0.140 is less than 5/4. C Oxygen, because 0.140/5 is less than 0.120/4. D Oxygen, because 0.140/4 is greater than 0.120/4.
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Compare 0.120/4=0.0300 mol with 0.140/5=0.0280 mol. Oxygen permits the smaller reaction amount and limits the modelled conversion.

Final answer

C

Question B4

[1 mark]
A stoichiometric feed contains 5.80\,\text{g} of reacting \mathrm{Na}_{2}\mathrm{S}\mathrm{O}_{3} and \mathrm{H}\mathrm{Cl} in total; solvent water is excluded. The supplied reaction is: \mathrm{Na}_{2}\mathrm{S}\mathrm{O}_{3}\mathrm{(s)}+2\,\mathrm{H}\mathrm{Cl}\mathrm{(aq)}\rightarrow2\,\mathrm{Na}\mathrm{Cl}\mathrm{(aq)}+\mathrm{S}\mathrm{O}_{2}\mathrm{(g)}+\mathrm{H}_{2}\mathrm{O}\mathrm{(l)} Sulfur dioxide is isolated in 84.0\% yield. What mass is isolated? Use section 7 of the data booklet.
A 0.784\,\text{g} B 1.57\,\text{g} C 1.87\,\text{g} D 2.22\,\text{g}
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M(Na2SO3)=126.05 and M(HCl)=36.46 g mol−1. A stoichiometric reaction batch uses 126.05+2(36.46)=198.97 g to form 64.07 g SO2. The isolated mass is 5.80×64.07/198.97×0.840=1.57 g.

Final answer

B; 1.57 g

Question B5

[1 mark]
Two batches use the same balanced equation, the same pure reactant amounts and the same desired product. One batch loses some product during transfer. Which statement is correct?
A The batch with transfer loss has a lower percentage yield but the same atom economy. B Both percentage yield and atom economy are lower in the batch with transfer loss. C Both quantities are unchanged because the starting amounts are identical. D The batch with transfer loss has a higher percentage yield but the same atom economy.
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Product lost during transfer reduces the recovered amount. The balanced reaction and designated product stay the same, so atom economy is unchanged.

Final answer

A

Question B6

[1 mark]
For \ce{CaCO3 -> CaO + CO2}, a pure carbonate charge of mass a gives an isolated lime mass b. Their molar masses are M_c and M_o. Select the percentage-yield expression.
A 100bM_c/(aM_o) B 100aM_o/(bM_c) C 100b/a D 100ab/(M_cM_o)
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The equation gives one mole of lime per mole of carbonate. The possible lime amount is a/M_c; the recovered amount is b/M_o. Dividing recovered by possible and multiplying by 100 gives 100(b/M_o)/(a/M_c)=100bM_c/(aM_o), option A. B reverses the comparison, C ignores different molar masses, and D multiplies amounts.

Final answer

A; 100bM_c/(aM_o)

Question B7

[2 marks]
Give the balanced net ionic equation for precipitating strontium phosphate from \ce{Sr^{2+}(aq)} and \ce{PO4^{3-}(aq)}.
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A neutral formula unit needs three divalent strontium ions and two trivalent phosphate ions, so the solid is \ce{Sr3(PO4)2} [1]. The account is \ce{3Sr^{2+}(aq) + 2PO4^{3-}(aq) -> Sr3(PO4)2(s)} [1, including states]. Its input charge is 3(+2)+2(-3)=0, matching the neutral solid. Sr 3, P 2 and O 8 also close. Keep the precipitate intact rather than writing its ions as dissolved products.

Final answer

\ce{3Sr^{2+}(aq) + 2PO4^{3-}(aq) -> Sr3(PO4)2(s)}

Question B8

[2 marks]
A 15.00\,\mathrm{cm^3} portion of 0.1600\,\mathrm{mol\,dm^{-3}} HCl is neutralised by 0.1200\,\mathrm{mol\,dm^{-3}} NaOH. Find the alkali volume at equivalence.
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The reacting equation is \ce{HCl + NaOH -> NaCl + H2O}. Acid amount =0.1600\times0.01500=0.002400 mol [1]. Equal alkali amount is needed, so V=0.002400/0.1200=0.02000 dm^3=20.00 cm^3 [1].

Final answer

20.00\,\mathrm{cm^3}

Tutorial practice

Question U1

[4 marks]
In a nitrogen-oxide scrubber, \mathrm{NH_{3}}\mathrm{(g)} and \mathrm{NO}\mathrm{(g)} react with oxygen to give only \mathrm{N_{2}}\mathrm{(g)} and water vapour.
(a) [ 2 ] {Write the balanced equation with the smallest integer coefficients and all states.}
(b) [ 2 ] {For 0.0800\,\mathrm{mol} \mathrm{NO} removed, determine the amounts of ammonia consumed and nitrogen formed.}
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Balance nitrogen and hydrogen together, then oxygen: the coefficients 4:4:1:4:6 conserve eight N, twelve H and six O atoms [1], with all gaseous states [1]. NO, NH3 and N2 each have coefficient four, giving 0.0800 mol consumed NH3 [1] and formed N2 [1].

Final answer

(a) 4NH3(g)+4NO(g)+O2(g)→4N2(g)+6H2O(g); (b) 0.0800 mol NH3 and 0.0800 mol N2

Question U2

[5 marks]
A hydrate \mathrm{Na_{2}CO_{3}} \cdot x\,\mathrm{H_2O} is analysed through its carbonate content. A 2.8619\,\mathrm{g} sample is dissolved and made up to 100.00\,\mathrm{cm^3}. A 20.00\,\mathrm{cm^3} aliquot consumes 20.00\,\mathrm{cm^3} of 0.2000\,\mathrm{mol\,dm^{-3}} \mathrm{HCl}. Each carbonate ion consumes two hydrogen ions.
(a) [ 2 ] {Calculate the hydrate amount in the original sample.}
(b) [ 3 ] {Calculate the hydrate molar mass and deduce x.}
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Acid delivered is 0.2000×0.02000=0.004000 mol; the aliquot contains half as much carbonate [1]. Multiply by 100.00/20.00 to obtain 0.01000 mol hydrate [1]. Divide 2.8619 g by this amount: 286.19 g mol−1, printed 286.2 [1]. Na2CO3 contributes 105.99 g mol−1, leaving 180.20 g mol−1 for water [1]. Divide that difference by 18.02: x=10 [1].

Final answer

(a) 0.01000 mol; (b) 286.2 g mol−1, x=10

Question U3

[4 marks]
A high-temperature process follows \mathrm{SiO_2}+3\,\mathrm{C}\rightarrow\mathrm{SiC}+2\,\mathrm{CO}. The initial amounts are 0.0500\,\mathrm{mol} silica and 0.180\,\mathrm{mol} carbon.
(a) [ 2 ] {Identify the limiting reactant, with a numerical comparison.}
(b) [ 2 ] {Determine the amounts of silicon carbide formed and carbon left after completion.}
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Available reaction amounts are 0.0500/1=0.0500 mol and 0.180/3=0.0600 mol [1]. SiO2 permits the smaller conversion [1]. SiC formed equals 0.0500 mol [1]. Carbon consumption is 3×0.0500=0.150 mol; subtract from 0.180 mol to leave 0.030 mol [1; retain decimal-place precision].

Final answer

(a) SiO2 limits; (b) 0.0500 mol SiC, 0.030 mol C remains

Question U4

[5 marks]
A pigment batch follows 2\,\mathrm{Cr(OH)_3(s)}\rightarrow\mathrm{Cr_2O_3(s)}+3\,\mathrm{H_2O(g)}. Starting with 10.30\,\mathrm{g} pure hydroxide, 6.08\,\mathrm{g} pure dry oxide is isolated.
(a) [ 3 ] {Calculate the theoretical oxide mass.}
(b) [ 2 ] {Calculate the percentage yield and give one possible reason it is below 100\,\%.}
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M(Cr(OH)3)=103.03 g mol−1 [1]. The oxide amount is (10.30/103.03)/2=0.04998544 mol [1]. Multiply by 152.00 to obtain 7.597786… g, printed 7.598 [1]. Yield=100×6.08/7.597786…=80.0% [1]. Material left on a vessel lowers the mass recovered [1].

Final answer

(a) 7.598 g; (b) 80.0%, for example transfer loss

Question U5

[4 marks]
A heat-storage cycle releases oxygen by 2\,\mathrm{BaO_2(s)}\rightarrow2\,\mathrm{BaO(s)}+\mathrm{O_2(g)}. Oxygen is the desired product.
(a) [ 2 ] {Calculate the oxygen atom economy.}
(b) [ 2 ] {Explain why a 100\,\% experimental oxygen yield would not make the oxygen atom economy 100\,\%.}
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The reactant side accounts for 2×169.33=338.66 g per written reaction [1]. Only 32.00 g is desired oxygen: 100×32.00/338.66=9.45% [1]. Full oxygen recovery is a statement about experimental yield [1]; the barium oxide still contains most of the input mass, so oxygen atom economy remains 9.45% [1].

Final answer

(a) 9.45%; (b) yield leaves the equation-based atom fraction unchanged

Question U6

[4 marks]
An antacid has 0.1750\,\mathrm{g} pure \ce{Mg(OH)2} per tablet; other ingredients do not consume acid. Use M=58.33\,\mathrm{g\,mol^{-1}}. (a) [ 2 ] {Determine the maximum amount of \ce{H+} neutralised by one tablet.} (b) [ 2 ] {Find the volume of 0.08000\,\mathrm{mol\,dm^{-3}} monoprotic acid with this capacity.}
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Each hydroxide formula unit consumes two protons: \ce{Mg(OH)2 + 2H+ -> Mg^2+ + 2H2O}. (a) n=0.1750/58.33=0.0030001714 mol hydroxide [1]; capacity =2n=0.0060003429 mol, or 0.006000 mol [1]. (b) V=0.0060003429/0.08000=0.075004286 dm^3 [1], hence 75.00 cm^3 to four significant figures [1]. The amount is a theoretical capacity, assuming complete reaction.

Final answer

0.006000\,\mathrm{mol}; 75.00\,\mathrm{cm^3}

Question U7

[5 marks]
A 0.6000\,\mathrm{g} powder contains \ce{CaCO3} and acid-inert filler. It receives 30.00\,\mathrm{cm^3} of 0.3000\,\mathrm{mol\,dm^{-3}} HCl. After reaction, all remaining acid needs 12.50\,\mathrm{cm^3} of 0.2000\,\mathrm{mol\,dm^{-3}} NaOH. (a) [ 2 ] {Find the amount of acid used by the carbonate.} (b) [ 2 ] {Determine the carbonate mass percentage.} (c) [ 1 ] {Explain why reaction must finish before titrating the remainder.}
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(a) Acid added =0.03000(0.3000)=0.009000 mol; remaining acid =0.01250(0.2000)=0.002500 mol [1]. Used acid =0.009000-0.002500=0.006500 mol [1]. (b) \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} gives carbonate amount 0.003250 mol [1]. Its mass is 0.003250(100.09)=0.3252925 g; purity =100(0.3252925)/0.6000=54.2154167\%=54.22\% [1]. (c) An unfinished reaction leaves too much acid; subtracting this remainder underestimates the carbonate content [1].

Final answer

0.006500\,\mathrm{mol} HCl used; 54.22\% \ce{CaCO3}; otherwise unreacted carbonate is undercounted.

Question U8

[5 marks]
A 1.200\,\mathrm{g} sample contains KOH and alkali-inert solid. All KOH is extracted and made to 250.0\,\mathrm{cm^3}. A 25.00\,\mathrm{cm^3} portion needs 16.00\,\mathrm{cm^3} of 0.1000\,\mathrm{mol\,dm^{-3}} HCl. (a) [ 3 ] {Calculate the sample's KOH mass percentage.} (b) [ 2 ] {The volumetric flask is overfilled. Predict the error in the calculated percentage if 250.0\,\mathrm{cm^3} is still used. Explain.}
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(a) \ce{KOH + HCl -> KCl + H2O} is 1:1. The aliquot contains 0.01600(0.1000)=0.001600 mol KOH [1]. The whole flask contains ten times this, 0.01600 mol [1]. With M(\ce{KOH})=39.10+16.00+1.01=56.11 g mol^{-1}, mass =0.89776 g, giving 100(0.89776)/1.200=74.8133333\%=74.81\% [1]. (b) Overfilling lowers concentration, so a fixed 25.00 cm^3 aliquot consumes less acid [1]. Scaling its result by the assumed factor ten misses the extra flask volume and gives too small a KOH amount and purity [1].

Final answer

74.81\%; too low because the portion contains less KOH than assumed.

Advanced practice

Question A1

[1 mark]
An alkane undergoes complete combustion. Its general formula is \mathrm{C}_n\mathrm{H}_{2n+2}. The products are carbon dioxide and water. With the fuel coefficient set to 2, what is the sum of coefficients in the balanced equation?
A 7n+5 B 7n+3 C 6n+5 D \dfrac{7n+5}{2}
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With fuel coefficient two, put 2n before CO2 and 2n+2 before H2O. Product oxygen totals 6n+2 atoms, requiring 3n+1 O2. Adding the four coefficients gives 2+(3n+1)+2n+(2n+2)=7n+5. This symbolic count is common content, not AHL.

Final answer

A; 7n+5

Question A2

[5 marks]
In a gas-yield test, 0.0120\,\mathrm{mol} of aluminium reacts with 0.0250\,\mathrm{mol} of hydrogen ions according to:
2\,\mathrm{Al(s)}+6\,\mathrm{H^+(aq)}\rightarrow2\,\mathrm{Al^{3+}(aq)}+3\,\mathrm{H_2(g)}.
Use R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}.
(a) [ 2 ] {Identify the limiting reactant using a calculation.}
(b) [ 2 ] {Determine the theoretical dry gas volume at 298\,\mathrm{K} and 1.00\times10^5\,\mathrm{Pa}.}
(c) [ 1 ] {The measured dry volume under these conditions is 2.70\times10^{-4}\,\mathrm{m^3}. Calculate the percentage yield.}
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Al would need 3×0.0120=0.0360 mol H+ [1]; only 0.0250 mol is supplied, so acid limits [1]. H2 amount is (3/6)×0.0250=0.0125 mol [1]. Ideal-gas calculation gives 0.0125×8.31×298/100000=0.0003095475 m3, printed 3.10×10−4 m3 [1]. Yield=100×0.000270/0.0003095475=87.2% [1]. Connect S1.5; dry volumes and matching conditions are essential.

Final answer

(a) H+; (b) 3.10×10−4 m3; (c) 87.2%

Question A3

[5 marks]
A manufacturer compares two routes to the same pure product. Both equations are balanced. Route A: \mathrm{C_4H_8}+\mathrm{H_2O}\rightarrow\mathrm{C_4H_{10}O}. Route B: \mathrm{C_4H_9Br}+\mathrm{NaOH}\rightarrow\mathrm{C_4H_{10}O}+\mathrm{NaBr}. The product is the alcohol. Each route has a 90.0\,\% isolated yield.
(a) [ 3 ] {Calculate each route’s atom economy.}
(b) [ 2 ] {Explain which route has less stoichiometric by-product waste, and identify one other factor needed to judge the environmental impact.}
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Route A puts all input atoms in alcohol, giving 100% [1]. For B, the relevant molar masses are 137.03, 40.00 and 74.14 g mol−1 [1]. Calculate 100×74.14/(137.03+40.00)=41.9% [1]. A therefore makes less equation-defined by-product per alcohol amount [1]. Energy demand or solvent impacts need a separate comparison [1]; equal 90.0% yields do not resolve that comparison. Connect R3.4 through the supplied routes; mechanisms are not required.

Final answer

(a) A: 100%; B: 41.9%; (b) A has less stoichiometric waste; compare energy or solvent burden

Question A4

[7 marks]
A 1.000\,\mathrm{g} solid contains only \ce{MgCO3} and \ce{CaCO3}. Excess acid releases 0.01050\,\mathrm{mol} dry \ce{CO2} quantitatively. Use molar masses 84.32 and 100.09\,\mathrm{g\,mol^{-1}}, respectively. (a) [ 4 ] {Determine the mass percentage of \ce{MgCO3} in the solid.} (b) [ 2 ] {Find the dry gas volume at 298\,\mathrm{K} and 99.0\,\mathrm{kPa}.} (c) [ 1 ] {If gas escapes before measurement, predict the error in the inferred magnesium-carbonate percentage.}
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(a) Each carbonate yields one \ce{CO2}. Let x and y be their amounts in mol. Then x+y=0.01050 [1] and 84.32x+100.09y=1.000 [1]. Substitution gives x=(100.09(0.01050)-1.000)/(100.09-84.32)=0.00323050095 mol [1]. Magnesium carbonate mass =0.272395840 g, hence 27.2395840\%, reported as 27.24\% to four significant figures [1]. (b) V=nRT/P=0.01050(8.31)(298)/99000=0.000262644545 m^3 [1], or 0.263 dm^3 [1]. (c) At fixed sample mass, magnesium carbonate gives more gas per gram. A low measured gas amount therefore gives too low a magnesium-carbonate percentage [1].

Final answer

27.24\% \ce{MgCO3}; 0.263\,\mathrm{dm^3}; too low.

Connected questions

Question C1

Connected practice[7 marks]
R2.1 + S1.5 + R1.1. A 0.5600\,\mathrm{g} iron hand-warmer reacts completely by \ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}. For this exercise the reaction enthalpy is -1650\,\mathrm{kJ} per equation as written. (a) [ 2 ] {Find the oxygen volume consumed at 298\,\mathrm{K} and 100\,\mathrm{kPa} (three significant figures).} (b) [ 3 ] {All released heat warms 50.0\,\mathrm{g} water, initially at 20.0\,{}^\circ\mathrm{C}. Find its final temperature. Use c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}.} (c) [ 2 ] {Explain the sign of the reaction enthalpy and how heat loss changes the water temperature.}
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(a) n(\ce{Fe})=0.5600/55.85=0.0100268577 mol; oxygen amount =3n/4=0.00752014324 mol [1]. V=nRT/P=0.00752014324(8.31)(298)/100000=0.000186227323 m^3=0.186 dm^3 [1]. (b) Equation extent =n(\ce{Fe})/4=0.00250671441 mol [1]. Heat released =1650(0.00250671441)=4.13607878 kJ [1]. Water rise =4136.07878/(50.0(4.18))=19.7898506 K, giving 39.8\,{}^\circ\mathrm{C} [1]. (c) Negative enthalpy means the reacting system releases energy to its surroundings [1]. Heat escaping to the air reduces the energy entering the water and lowers its final temperature [1].

Final answer

0.186\,\mathrm{dm^3} \ce{O2}; 39.8\,{}^\circ\mathrm{C}; exothermic, heat loss lowers the temperature.

Question C2

Connected practice[7 marks]
R2.1 + R3.1. Mix 30.00\,\mathrm{cm^3} of 0.08000\,\mathrm{mol\,dm^{-3}} HCl with 20.00\,\mathrm{cm^3} of 0.1000\,\mathrm{mol\,dm^{-3}} NaOH. Both are strong; assume complete dissociation, additive volumes and 25\,{}^\circ\mathrm{C}. (a) [ 3 ] {Write the net ionic equation and calculate the amount of excess reacting ion.} (b) [ 2 ] {Calculate the final pH.} (c) [ 2 ] {Find the extra volume of the same NaOH needed for equivalence and explain why adding water alone cannot reach equivalence.}
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(a) \ce{H+(aq) + OH-(aq) -> H2O(l)} [1]. Initial acid amount =0.03000(0.08000)=0.002400 mol and hydroxide =0.02000(0.1000)=0.002000 mol [1]. Hydroxide limits, leaving 0.000400 mol \ce{H+} [1]. (b) Total volume =0.05000 dm^3, so [\ce{H+}]=0.000400/0.05000=0.00800 mol dm^{-3} [1]. \mathrm{pH}=-\log_{10}(0.00800)=2.09691=2.10 [1]. (c) Extra alkali volume =0.000400/0.1000=0.00400 dm^3=4.00 cm^3 [1]. Water lowers concentration but adds no hydroxide amount to consume the excess acid [1].

Final answer

\ce{H+ + OH- -> H2O}; 0.000400\,\mathrm{mol} \ce{H+}; pH 2.10; 4.00\,\mathrm{cm^3} NaOH.

Question C3

Connected practice[7 marks]
R2.1 + R3.4 + S3.2. Two supplied routes give ethanol: A, \ce{C2H4 + H2O -> C2H6O}; B, \ce{C2H5Br + NaOH -> C2H6O + NaBr}. Both recover 0.1000\,\mathrm{mol} ethanol; isolated yields are 80.0\% for A and 90.0\% for B. Feeds have exact equation ratios. (a) [ 2 ] {Name the reaction class for each route and identify ethanol's characteristic functional group.} (b) [ 2 ] {Calculate each route's atom economy.} (c) [ 3 ] {Determine the total reacting feed mass needed for each route and explain why the higher-yield route can still require more feed.}
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(a) A adds water across the alkene double bond; B substitutes hydroxyl for bromine [1]. Ethanol contains the hydroxyl group, \ce{-OH} [1]. (b) A retains all reactant atoms in the desired product, giving 100% [1]. B has reacting molar mass M(\ce{C2H5Br})+M(\ce{NaOH})=108.97+40.00=148.97 g mol^{-1} and economy 100(46.08)/148.97=30.9324025\%=30.9\% [1]. (c) A needs 0.1000/0.800=0.1250 mol of equation batches, so feed mass =0.1250(46.08)=5.76 g [1]. B needs 0.1000/0.900=0.111111111 mol of batches, giving 0.111111111(148.97)=16.6 g [1]. Although B recovers a larger fraction of its possible ethanol, much of its stoichiometric feed becomes sodium bromide rather than ethanol [1].

Final answer

Addition; substitution; hydroxyl. Atom economies A 100\%, B 30.9\%. Feed masses A 5.76\,\mathrm{g}, B 16.6\,\mathrm{g}.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

How much? The amount of chemical change from the core ideas to full exam questions, with question types, worked examples and graded practice.

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