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.

Step-by-step 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* 18.0L=n_(CO_2)* 0.0821\text{ L atm }mol^(-1)K^(-1)* 593K\\\\n_(CO_2)=(1.00* 18.0)/(0.0821* 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 =
`(1)/(1)* 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* 100g/mol)=37g`

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