Question

25 gram of calcium carbonate completely reacts with sulphuric acid to produce 11 gram of carbon dioxide the percentage purity of calcium carbonate will be

Answer 1

the CaCO3 is pure... so the percentage purity is 100%

Answer 2

Molecular weight of $CaCO _{3}$=100g

Molecular weight of $CO _{2}$=44g

Decomposition of $CaCO _{3}$⟶$CaO+CO_{2}$

​Now, from the above reaction-Weight of pure $CaCO _{3}$required to produce 44g of $CO _{2}$=100g

Weight of pure required to produce 11g of $CO_{2}$= $\frac{100}{44}$*11≈25g

Given weight of $CaCO _{3}$ =25g

Now,

Amount of pure $CaCO_{3}$ in 25g of given $MgCO_{3}$ =25g

Thus,

Amount of pure $CaCO _{3}$ in 100g of given $MgCO_{3}$= $\frac{25}{25}$×100=100g

Therefore,

The percentage purity of given $CaCO_{3}$

​ =100%

#spj3