Answer to Question #145719 in General Chemistry for Vitaly

Question

A student conducted the following reaction in the lab: CaCO3= CaO + CO2
She began with 0.410 mol of CaCO3.
How many moles of CO2 should she expect to collect?
How many grams of CO2 should she expect to collect?
Suppose that she actually collected 16.80 g of CO2. What is her percent yield?

Answer 1

Solution:

CaCO₃ = CaO + CO₂

According to the reaction equation from 1 mol of CaCO₃ be produced 1 mole of CO₂.

molar mass of CO₂ 44.01 gram/mol

Therefore,

n of CO₂ = 0.410 moles

m of CO₂ = 0.410mol×44.01g/mol = 18.04 g

Percent yield = Actual yield/Theoretical yield × 100%

Percent yield = 16.80/18.04× 100% = 93.13%

Answer: n of CO₂ = 0.410 moles

m of CO₂ = 18.04 g

Percent yield = 93.13%