Question

In the reaction, MgCO,(s) MgO(s)+CO,(g) what mass of magnesium carbonate, MgCO, would be required to produce 183 L of carbon dioxide, CO2, measured at STP? mass:

Answer 1

Molar volume of any gas at STP = 22.4 L/mol

So,

Mol of CO2 produced = volume /molar volume

= 183 L / 22.4 L/mol

= 8.17 mol

From reaction,

Mol of MgCO3 reacted = mol of CO2 produced

= 8.17 mol

Molar mass of MgCO3,

MM = 1*MM(Mg) + 1*MM(C) + 3*MM(O)

= 1*24.31 + 1*12.01 + 3*16.0

= 84.32 g/mol

use:

mass of MgCO3,

m = number of mol * molar mass

= 8.17 mol * 84.32 g/mol

= 689 g

Answer: 689 g