**1. **Sulfur dioxide is an important chemical used in the manufacture of sulfuric acid by a process called the Contact process. Industrially, there are two methods for producing sulfur dioxide.

__Method 1__: burning sulfur in air S(l) + O_{2}(g) → SO_{2}(g)

__Method 2__: roasting iron sulfide in air _{2}(s) + 11O_{2}(g) → 2Fe_{2}O_{3}(s) + 8SO_{2}(g)

a) Explain why the reaction for method 1 has a percentage atom economy of 100%.

*Answer:* There is only one product / no undesired product, so none of the reactants are wasted

b) Calculate the percentage atom economy for producing sulfur dioxide by method 2.

Relative formula masses (M_{r}): FeS_{2} = 120 ; O_{2} = 32 ; _{2}O_{3} = 160 ;_{2} = 64

*Answer:* Atom economy = 100 × (sum of the M_{r} of desired products ÷ sum of the M_{r} of all the reactants)

c) 100 kg of iron sulfide (FeS_{2} ) were roasted in an excess of air. Calculate the maximum theoretical yield of sulfur dioxide that this mass of FeS_{2} can produce. Relative formula masses (M_{r}): FeS_{2} = 120 ; _{2} = 64

*Answer:* mol FeS_{2} = mass ÷ M_{r} = 100 000 ÷ 120 = 833.3 mol

mol SO_{2} = 2 × mol FeS_{2} (from the 4 : 8 ratio in the balanced equation) =1666.7 mol SO_{2}

mass SO_{2} = mol × M_{r} = 1666.7 × 64 = 106666 g = 106.7 kg

d) The actual yield of sulfur dioxide in method 2 is 95.0 kg. Calculate the percentage yield for this reaction.

*Answer:* % yield = 100 × (actual yield max. theoretical yield)

**2. **A student carried out a titration in order to determine the concentration of sulfuric acid in a solution. They titrated 25.0 cm^{3} portions of the sulfuric acid solution with a 0.850 mol/dm^{3} solution of potassium hydroxide. The equation for the reaction is: H_{2}SO_{4} + 2KOH → K_{2}SO_{4} + 2H_{2}O. The student’s results are shown in the table below.

a) Calculate the mean value for the volume of potassium hydroxide solution added. Give your answer to two decimal places. Write down the uncertainty in these three titration volumes.

*Answer:* mean volume = (23.55 + 23.40 + 23.45) ÷ 3 ^{3} ± 0.08 cm^{3}

b) Use the answer and the information in the table above to calculate the concentration of the sulfuric acid solution in mol/dm^{3}.

*Answer:* mol KOH = concentration × volume = ^{3} × (23.47 ÷ 1000) = 0.01995 mol ;

mol H_{2}SO_{4} = 12 × mol KOH (from 1 : 2 ratio in balanced equation) = 0.009975 mol ;

concentration of H_{2}SO_{4} = mol ÷ volume ^{3}

c) Use the answer above to calculate the mass of sulfuric acid dissolved in 25.0 cm^{3} of the sulfuric acid solution. Relative formula mass (M_{r}) H_{2}SO_{4} = 98

*Answer:* mol H_{2}SO_{4} in 25.0 cm^{3} = 0.009975 mol ;

mass H_{2}SO_{4} = mol × M_{r} = 0.009975 × 98 = 0.978 g

**3. **Urea has the formula CH_{4}N_{2}O, and is an important chemical used in agriculture, the chemical industry and in the manufacture of explosives. Urea can be manufactured by reacting ammonia with carbon dioxide: _{3}(g) + CO_{2}(g) → CH_{4}N_{2}O(g) + H_{2}O(g)

a) Give the number of moles of urea that will form when 0.46 mol of ammonia gas are reacted with an excess of carbon dioxide gas.

*Answer:* Volume ratio of NH_{3} : CH_{4}N_{2}O = 2 : 1 ; _{4}N_{2}O = ^{1}⁄_{2} × mol NH_{3} = 0.23 mol

b) In one reaction, 5.00 g of ammonia reacts with 5.00 g of carbon dioxide. Show that the carbon dioxide is the limiting reactant in this reaction. Relative formula masses (M_{r}): NH_{3} = 17 ; CO_{2} = 44

*Answer:* mol NH_{3} = mass ÷ M_{r} = 5.00 g ÷ 17

mol CO_{2} = mass ÷ M_{r} = 5.00 g ÷ 44 = 0.114 mol ;

so, 0.294 mol of NH_{3} will react with 0.147 mol of CO_{2} . There are fewer moles of CO_{2} , so CO_{2} will run out before the NH_{3} . CO_{2} is the limiting reactant.

c) Urea is highly soluble in water. A solution of urea was prepared by dissolving 0.75 g of urea into 400 cm^{3} of water. Calculate the concentration of this urea solution in g/dm^{3}.

*Answer:* concentration = mass ÷ volume ^{3}

d) Convert the concentration of urea in the answer above from g/dm^{3} into mol/dm^{3}. Relative formula _{r}) CH_{4}N_{2}O = 70

*Answer:* mol/dm^{3} = g/dm^{3} ÷ M_{r} = 1.88 ÷ 70 ^{3}

**4. **When heated, ammonium dichromate [(NH_{4})_{2}Cr_{2}O_{7} ] decomposes to form chromium oxide (Cr_{2}O_{3} ), nitrogen gas (N_{2} ) and water. A student heated 7.56 g of ammonium dichromate until it had completely decomposed. 4.56 g of chromium oxide, 0.84 g of nitrogen gas, and 2.16 g of water were formed.

a) Calculate the relative formula mass of (NH_{4})_{2}Cr_{2}O_{7} . Relative atomic masses (A_{r}): N = 14 ;

*Answer:* M_{r} (NH_{4})_{2}Cr_{2}O_{7} = (2 × 14) + (8 × 1) + (2 × 52)

b) Use the answer and the information above, to determine the balanced equation for this reaction. Relative formula masses (M_{r}): Cr_{2}O_{3} = 152 ; N_{2} = 28 ; H_{2}O = 18

*Answer:* mol (NH_{4})_{2}Cr_{2}O_{7} = mass ÷ Mr = 7.56 ÷ 252 = 0.03 mol ;

mol Cr_{2}O_{3} = 4.56 ÷ 152 = 0.03 mol ;

mol N_{2} = 0.84 ÷ 28 = 0.03 mol ;

mol H_{2}O = 2.16 ÷ 18 = 0.12 mol

Dividing each number of moles by 0.03 mol to give the whole number ratio. _{4})_{2}Cr_{2}O_{7} : Cr_{2}O_{3} : N_{2} : H_{2}O = 1 : 1 : 1 : 4

Correct symbol equation: _{4})_{2}Cr_{2}O_{7} → Cr_{2}O_{3} + N_{2} + 4H_{2}O