Question

In the following reaction, what mass of magnesium carbonate, MgCO3, would be required to produce 673 L of carbon dioxide, CO2, measured at STP?

Answer 1

You haven’t mentioned the reaction, but I would use the most common reaction between MgCO₃ and HCl to answer the question.

MgCO₃ + 2 HCl -------> MgCl₂ + H₂O + CO₂

The molar ratio of MgCO₃ and HCl is 1:1. We will assume CO₂ to behave as an ideal gas.

We know, from the gas laws that 1 mole of an ideal gas has a volume of 22.4 L = 22400 mL at STP.

Therefore, moles of CO₂ produced at STP = (673 L)*(1 mole/22.4 L) = 30.0446 moles (I will keep a few guard digits).

Molar mass of MgCO₃ = 84.3139 g/mol.

As per the balanced equation, 1 mole CO₂ is obtained from 84.3139 g MgCO₃.

Therefore, 30.0446 mole CO₂ is produced from (30.0446 mole CO₂)*(1 mole MgCO₃/1 mole CO₂)*(84.3139 g MgCO₃/1 mole MgCO₃) = 2533.1774 g MgCO₃ = (2533.1774 g)*(1 kg/1000 g) = 2.5331 kg MgCO₃ ≈ 2.53 kg MgCO₃ (ans).