Chemical Formulae, Equations and Calculations Part 1
3 marks · EasyA student reacts 8.0 g of hydrogen with excess oxygen to make water, according to the equation:
The student obtains 70.0 g of water. (Ar: H = 1, O = 16)
Calculate the maximum mass of water that could be formed from 8.0 g of hydrogen.
Calculate the percentage yield of water.
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(a) 72 g · must show Correct calculation of moles of H2 (4 mol), Stoichiometry from balanced equation, Calculation of theoretical mass of H2O (72g)
(b) 97% · must show Percentage yield = (actual/theoretical) × 100, Correct use of 70.0g and their answer from (a)
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Marking points
M1 (a) · 1 mark · Calculates number of moles of H2 from mass (8.0 / 2 = 4 mol). (follow-through)
A1 (a) · 1 mark · Uses correct stoichiometric relationship and calculates 72 g H2O as the theoretical maximum. (follow-through)
A2 (b) · 1 mark · Calculates percentage yield using correct numbers and formula, answer 97%. (follow-through)
Worked solution
a) Mr of H2 = 2 × 1 = 2. Number of moles of H2 = 8.0 / 2 = 4 mol.
From the balanced equation: 2 mol H2 make 2 mol H2O. So 4 mol H2 will form 4 mol H2O.
Mr of H2O = 2 × 1 + 16 = 18. Mass of H2O = 4 × 18 = 72 g.
b) Percentage yield = (actual / theoretical) × 100 = (70.0 / 72) × 100 = 97%.