Question

Quicklime (CaO) can be prepared by roasting

limestone (CaCO3) according to the reaction
CaCO3(s) ∆−→CaO(s) + CO2(g). When 3.3 × 103 g of CaCO3 are heated, the actual yield of CaO is 1 × 103 g. What is the percent yield?
Answer in units of %.

Answer 1

Answer: The percent yield is of CaO is 56%.

Step-by-step explanation:

The first solution provided by meguelratatouille came close to the correct answer, but erred in assuming that equal moles implies equal mass. The following discussion corrects this mistake.

To determine the percent yield, we must compare the actual yield to the theoretical yield.

Theoretical Yield

From the balanced chemical equation we have:

CaCO3(s) --> CaO(s) + CO2(g)

It is necessary to know the molar mass of CaCO3 and CaO to determine the amount of CaO produced from a given amount of CaCO3:

Molar mass CaCO3 = 100.09 g/mol

Molar mass CaO = 56.08 g/mol

Then, to find the total mass of CaO produced by 3.3 x 10^3 g of CaCO3 reactant, we calculate:

3.3 x 10^3 g CaCO3
x 1 mol CaCO3 / (100.09 g/mol CaCO3)
x 1 mol CaO / (1 mol CaCO3)
x 56.08 g CaO / (1 mol CaO)
-----------------------------------
1848 g CaO

which we report as 1.8 x 10^3 g CaO after applying significant figures.

Actual Yield

Finally, the percent yield is given by dividing the actual yield by the theoretical yield we just computed:

(1 x 10^3 g) / (1.8 x 10^3 g) x 100% = 55.6% = 56%

Answer 2

Step-by-step explanation:

To calculate the percent yield, we need to compare the actual yield to the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction went to completion based on the stoichiometry of the balanced equation.

First, let's determine the molar mass of CaCO3 (limestone) and CaO (quicklime):

- Molar mass of CaCO3 = 40.08 g/mol (molar mass of Ca) + 12.01 g/mol (molar mass of C) + (3 * 16.00 g/mol) (molar mass of O) = 100.09 g/mol

- Molar mass of CaO = 40.08 g/mol (molar mass of Ca) + 16.00 g/mol (molar mass of O) = 56.08 g/mol

Next, we can calculate the theoretical yield of CaO:

The molar ratio between CaCO3 and CaO is 1:1 according to the balanced equation. Therefore, the mass of CaO produced is the same as the mass of CaCO3 used.

The theoretical yield of CaO is 3.3 × 10^3 g.

Now we can calculate the percent yield:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

Percent Yield = (1 × 10^3 g / 3.3 × 10^3 g) * 100

Percent Yield = 30.30%

Therefore, the percent yield of CaO in this reaction is 30.30%.