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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.