Answer:
a.
C₂H₂ (g) + 5/2 O₂ (g) ⇒ 2CO₂ (g) + H₂O (l)
b. 6.21 g H₂O
c. 1.08 g
d. 6.21 g H₂O
e. 16 %
Explanation:
This question involves a calculation based on the stoichiometry of the balanced chemical equation:
b.
Lets calculate the # moles C₂H₂ 8.98 g will represent and then calculate the amount of water produced as follows:
# moles C₂H₂ = mass/molar mass = 8.98 g / 26.04 g/mol = 0.34 mol
From the stoichiometry of the reaction:
1 mol H₂O produced / mol C₂H₂ x 0.34 mol C₂H₂ = 0.34 mol H₂O produced
g H₂O = # mol H₂O x molar mass H₂O = 0.34 mol x 18.01 g/mol = 6.21 g H₂O
c.
For 4.58 g O₂ we can calculate the amount of water in grams formed as follows:
# mol O₂ = mass / molar mass O₂ = 4.58 g / 32 g / mol = 0.14 mol
From the stoichiometry of the reaction we have
1 mol H₂O produced /2.5 mol O₂ x 0.14 mol O₂ = 0.06 mol H₂O
mass H₂O produced = 0.06 mol x molar mas H₂O = 0.06 mol x 18.01 g/mol
= 1.08 g H₂O
d,e.
We calculated in part b that we should have produced 6.21 g H₂O, therefore the percent yield is =
1 g / 6.21 g x 100 g = 16 %
Note one could argue that this theoretical yield refers to the 4.58 grams O₂ in part c. However if that were the case we will have more than 100 % yield, unless we round the numbers to give us 100 % yield
