Question

The mineral magnesite contains magnesium carbonate, MgCO3 (molar mass= 84 g/mol), and other impurities. When a 1.26-g sample of magnesite was dissolved in hydrochloric acid, 0.22 g of CO2 was generated. If the magnesite contained no carbonate other than MgCO3, what was the percent MgCO3 by mass in the magnesite? The reaction is as follows:
MgCO3+2HCl-->MgCl2+H2O+CO2

Answer 1

\[\rm{MgCO_3+2HCl\rightarrow{MgCl_2} + H_2O+CO_2}\]
This tells us that :

\[\rm{\text{1 mole}\space{CO_2}\text{is produced from 1mole}\ MgCO_3}\]
Converting moles to grams:

\[\rm{\text{So}\space{44g}\space{CO_2}\space{\text{is produced from}}\space{84g\space{MgCO_3}}}\]

\[\rm{\text{So}\space{1.0g}\space{CO_2}\space{\text{is produced from}}\space\frac{84}{44}g\space{MgCO_3}}\]
\[\rm\text{So}\space{0.22g}\space{CO_2}\text{ is produced from}\space{\frac{84}{44}\times{0.22}}g={0.42g}\space{MgCO_3}\]


So the percentage of magnesium carbonate in the sample is given by:
\[\boxed{\frac{0.42}{1.26}\times{100}=33\%}\]