When acetylene gas, C2H2, reacts with oxygen gas the products are carbon dioxide and water. a. What is the balanced equation? b. How many grams of water can be formed from the consumption of 8.98 grams of acetylene c. How many grams of water can be formed from the consumption of 4.58 grams of oxygen d. What is the theoretical yield of the reaction? e. The reaction yielded 1.00g of water. What is the percent yield?

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