If 3.22 g of precipitate are recovered from the reaction of limewater (Ca(OH),) with carbon dioxide to produce water and calcium carbonate, what is the percent yield if 20.0 g of carbon dioxide was mixed with the
limewater?
the correct answer is 7.1% but i don't understand what they did

Percent yield = 7.1%

Explanation:

The general reaction of the problem is:

Ca(OH)₂ + CO₂ → H₂O + CaCO₃

To solve this question we need to find theoretical yield using the amount of carbon dioxide added because:

Percent yield = Actual yield (3.22g) / Theoretical yield * 100

Theoretical yield is the maximum amount of product that could be obtained. To find it we need to convert the mass of CO₂ to moles. The moles of CO₂ = Moles of CaCO₃:

Moles CO₂ -Molar mass: 44.01g/mol):

20.0g * (1mol / 44.01g) = 0.454 moles CO₂ = Moles of CaCO₃ produced

Mass CaCO₃ = Theoretical yield -Molar mass: 100.09g/mol-:

0.454 moles * (100.09g / mol) = 45.5g of CaCO₃ = Theoretical yield.

Replacing:

Percent yield = 3.22g / 45.5g * 100

14,16 18 reacts with group

the answer is c. nacl