A mixture contained 96.4 g of CaCO₃ (molar mass of 100.09) and 68.3 g of MgCO₃ (molar mass of 84.31) . How many grams of CO₂ (molar mass of 44.01) will be produced when the mixture is heated?
CaCO₃(g) $\rightarrow$ CaO(s) + CO₂(g)
MgCO₃(g) $\rightarrow$ MgO(s) + CO₂(g)

Question

A mixture contained 96.4 g of CaCO₃ (molar mass of 100.09) and 68.3 g of MgCO₃ (molar mass of 84.31) . How many grams of CO₂ (molar mass of 44.01) will be produced when the mixture is heated?
CaCO₃(g)

\rightarrow

CaO(s) + CO₂(g)
MgCO₃(g)

\rightarrow

MgO(s) + CO₂(g)

Quizplus · Accepted Answer

The moles of CO₂ produced from CaCO₃ is 96.4 g / 100.09 g/mol = 0.963 moles. The moles of CO₂ produced from MgCO₃ is 68.3 g / 84.31 g/mol = 0.810 moles. Total moles of CO₂ = 0.963 + 0.810 = 1.773 moles. The mass of CO₂ produced is 1.773 moles × 44.01 g/mol = 78.0 g.

A Mixture Contained 96 →\rightarrow→ CaO(s) + CO2(g) MgCO3(g) →\rightarrow→ MgO(s) + CO2(g) A) 32

A Mixture Contained 96 $\rightarrow$ CaO(s) + CO₂(g)
MgCO₃(g) $\rightarrow$ MgO(s) + CO₂(g)
A) 32