Question

$3.88\, g$ of a mixture of $Na _{2} CO _{3}$ and $CaCO _{3}$ is heated to a constant mass of $3.00\, g$. What was the mass of $Na _{2} CO _{3}$ (in grams) in the mixture?

Answer 1

$Na _{2} CO _{3}$ does not decompose on heating. $CaCO _{3}$ decomposes to form $CaO$ and $CO _{2}$.
$CaCO _{3} \xrightarrow{\Delta} CaO + CO _{2}$
$\therefore $ Loss in mass is due to loss of $CO _{2}$ gas
$\therefore (3.88-3.00=0.88\, g )$ of $CO _{2}$ is lost during heating.
$0.88\, g\, CO _{2}=\frac{0.88}{44}=0.02\, mol\, CO _{2}$
From the reaction stoichiometry,
$0.02\, mol CO _{2} \equiv 0.02 \,mol\, CaCO _{3}$
$0.02\, mol\, CaCO _{3}=0.02 \times 100=2.00\, g\, CaCO _{3}$
$\therefore $ Mass of $Na _{2} CO _{3}$ in the mixture
$=3.88-2.00\, g =1.88\, g$