67.00 grams of calcium carbonate breaks down. a. write the balanced the reaction b. identify the reaction type (single replacement, double replacement, synthesis, decomposition or combustion). c. what is the theoretical yield of carbon dioxide gas (in grams)? d. what is the percent yield if 27.67 grams of carbon dioxide gas are actually produced?

A. CaCO₃(s) → CaO(s) + CO₂(g).

B. Decomposition reaction.

C. 29.46 g.

D. 93.94%

Explanation:

A. Write the balanced the reaction

Calcium carbonate breaks down according to the balanced equation:

CaCO₃(s) → CaO(s) + CO₂(g),

1.0 mol of CaCO₃ is decomposed to produce 1.0 mol of CaO and 1.0 mol of CO₂.

B. Identify the reaction type (single replacement, double replacement, synthesis, decomposition or combustion).

It is a decomposition reaction.

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds.

Herein, CaCO₃ is decomposed to produce CaO and CO₂.

C. What is the theoretical yield of carbon dioxide gas (in grams)?

Firstly, we should get the no. of moles (n) of 67.00 grams of calcium carbonate:

n = mass/molar mass = (67.0 g)/(100.08 g/mol) = 0.6695 mol.

Using cross-multiplication:

1.0 mol of CaCO₃ produces → 1.0 mol of CO₂, from stichiometry.

∴ 0.6695 mol of CaCO₃ produces → 0.6695 mol of CO₂.

We can calculate the theoretical yield of carbon dioxide gas using the relation:

∴ The theoretical yield of carbon dioxide gas = n*molar mass = (0.6695 mol)(44.0 g/mol) = 29.46 g.

D. What is the percent yield if 27.67 grams of carbon dioxide gas are actually produced?

∵ the percent yield = (actual yield/theoretical yield)*100.

actual yield = 27.67 g, theoretical yield = 29.46 g.

∴ the percent yield = (27.67 g/29.46 g)*100 = 93.94%.

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