Question

4 g of a sample of MgCO3 on strong heating gives 800 mL of CO2 at STP . calculate the percentage purity of the sample​

Answer 1

Answer ⤵️

MgCO

3

→MgO+CO

2

The molar masses of MgCO

3

and MgO are 24+12+48=84 g/mol and 24+16=40 g/mol respectively.

Thus, 8 g MgO will be obtained from

40

84

×8=16.8 mg of MgCO

3

.

The percentage purity of magnesium carbonate in the sample is

20

16.8

×100=80%

Hence, the correct option is B