Question

Solving for a gaseous reactant 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) 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO) into solid calcium oxide and gaseous carbon dioxide 2. Suppose 67.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 280.0 "Cand pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted. Round your answer to 3 significant digits.

Answer 1

1) Balanced chemical equation for given reaction is  CaCO _3(s)$\overset{\Delta }{\rightarrow}$ CaO _(s) + CO_2(g)

2) Given

1) Volume of CO₂ gas = 67.0 L

2) Temperature of a gas = 280.0 ⁰ C = 553.0 K

3) Pressure of a gas = 1 atm

From above data , we can calculate no. of moles of CO₂ produced .

We have relation, P V = n R T

where, P is a pressure of a gas, V is a volume of a gas, n is no. of moles of gas, R is a gas constant and T is temperature of gas.

$\therefore$ n = P V / R T = ( 1 atm $\times$ 67.0 L / 0.082057 L atm mol ^-1 K ^-1$\times$ 553.0 K )

n = 1.476 mol = no. of moles of CO₂ produced

From reaction, 1 mole CO₂ gas is produced from 1 mole CaCO ₃. Hence, 1.476 mol CO₂ gas is produced from 1.476 mole CaCO ₃ .

Hence, mass of CaCO ₃ reacted = No. of moles of CaCO ₃ reacted $\times$ Molar mass of CaCO ₃

$\therefore$ mass of CaCO ₃ reacted = 1.476 mol $\times$ ( 100 g / 1 mol ) = 147.6 g = 148 g

ANSWER :  mass of CaCO ₃ reacted = 148 g