Question

Ancient Romans built often out of bricks and mortar. A key ingredient in their mortar was quicklime (calcium oxide), which they produced by roasting limestone (calcium carbonate).Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO_3) into solid calcium oxide and gaseous carbon dioxide. CaCO3 → CaO + CO2 Suppose 18.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 320.0 degree C and pressure of exactly 1-atm. Calculate the mass calcium carbonate that must have reacted. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: The mass of calcium carbonate that must be reacted is 37 grams.

Explanation:

The balanced chemical equation for the decomposition of calcium carbonate follows:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

To calculate the moles of carbon dioxide gas, we use the equation given by ideal gas, which follows:

$PV=nRT$

where,

P = pressure of carbon dioxide = 1.00 atm

V = Volume of carbon dioxide = 18.0 L

T = Temperature of the mixture = $320^oC=[320+273]K=593K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of carbon dioxide = ?

Putting values in above equation, we get:

$1.00atm\times 18.0L=n_{CO_2}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 593K\\\\n_{CO_2}=\frac{1.00\times 18.0}{0.0821\times 593}=0.370mol$

By Stoichiometry of the reaction:

1 mole of carbon dioxide is produced from 1 mole of calcium carbonate

So, 0.370 moles of carbon dioxide will be produced from = $\frac{1}{1}\times 0.370=0.370mol$ of calcium carbonate

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = 0.370 moles

Putting values in above equation, we get:

$0.370mol=\frac{\text{Mass of calcium carbonate}}{100g/mol}\\\\\text{Mass of calcium carbonate}=(0.370mol\times 100g/mol)=37g$

Hence, the mass of calcium carbonate that must be reacted is 37 grams.