Question

A sample of 53.0 of carbon dioxide was obtained by heating 1.31 g of calcium carbont. What is the percent yield for this reaction?
CaCO3 ? CaO + CO2

Answer 1

Answer:

92.04%

Explanation:

Given:

Mass of CO₂ obtained = 53.0 grams

Mass of calcium carbonate heated = 1.31 grams

Now,

the molar mass of the calcium carbonate = 100.08 grams

The number of moles heated in the problem = Mass  / Molar mass

= (1.31 grams) / (100.08 grams/moles)

= 0.013088 moles

now,

1 mol of calcium carbonate yields 1 mol of CO₂

thus,

0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂

now,

Theoretical mass of 0.013088 moles of CO₂ will be

= Number of moles × Molar mass of CO₂

= 0.013088 × 44 = 0.5758  grams

Thus, the percent yield for this reaction = $\frac{\textup{Actual yield}}{\textup{Theoretical yield}}\times100$

or

the percent yield for this reaction = $\frac{0.53}{0.5758}\times100$

or

the percent yield for this reaction = 92.04%