The volume of ${CO}_2$ that can be obtained at STP from $60 g$. of $70\%$ pure of ${MgCO}_3$ is

The given reaction occurs as follows  ${MgCO}_3\to MgO+{CO}_2$

Acc. to the given reaction 84g${MgCO}_3$ produces 44g${CO}_2$ i.e 22.4L

Now here we are provided with 60g ${MgCO}_3$ which is 70 percent pure thus only 70 percent of it takes part in reaction hence the reactive part of given${MgCO}_3$= $60\times 0.70$= 42g

Now using unitary method, since 84g ${MgCO}_3$ produce 44g carbon dioxide

                          42g${MgCO}_3$ produces 22g ${CO}_2$ i.e 0.5 moles

0.5 mole further corresponds to 11.2L 

Hence option(B) is the correct answer.