Question

Calcium carbonate decomposes at high temperatures to form carbon dioxide and calcium oxide. What mass of CaCO₃ would be needed to produce 3.45L of CO₂ at 1.0 atm and 289K?

Answer 1

CaCO₃ → CO₂ + CaO

$V=3.45 \;L \\ p= 1.0 \;atm \\ T = 289 \;K$

Ideal gas law

pV =nRT

R = 0.08206 L×atm/mol×K

$n = \frac{pV}{RT} \\ n(CO_2) = \frac{1.0 \times 3.45}{0.08206 \times 289} = 0.145 \;mol$

According to the reaction:

n(CaCO₃) =n(CO₂) = 0.145 mol

MM(CaCO₃) = 100.08 g/mol

m(CaCO₃) $= 0.145 \times 100.08 = 14.56 \;g$

Answer: 14.56 g