Question

How many grams of $N{{a}_{2}}C{{O}_{3}}$ are produced when $1.5kg$ of $NaHC{{O}_{3}}$ is decomposed?

Answer 1

When a mixture contains $N{{a}_{2}}C{{O}_{3}}$ and $NaHC{{O}_{3}}$ , then on heating only $NaHC{{O}_{3}}$ produces the carbon dioxide not $N{{a}_{2}}C{{O}_{3}}$ . So write the decomposition reaction of $NaHC{{O}_{3}}$ and find the weight of $NaHC{{O}_{3}}$ required to produce litres of at STP and subtract this weight from grams to find the weight of $N{{a}_{2}}C{{O}_{3}}$ in the mixture before heating.

Complete answer:
The balanced chemical reaction is give as; $NaHC{{O}_{3(s)}}\to N{{a}_{2}}C{{O}_{3(s)}}+C{{O}_{2(g)}}+{{H}_{2}}{{O}_{\left( g \right)}}$
Firstly we need to convert $1.5kg$ into grams;
$1.5kg\times \dfrac{1000g}{1kg}=1500g$
Calculate mol $NaHC{{O}_{3}}$ by dividing it’s given mass by its molar mass $\left( 84.006\text{ }g/mol \right).$
We get; $1500g(NaHC{{O}_{3}})\times \dfrac{1mol(NaHC{{O}_{3}})}{84.006g(NaHC{{O}_{3}})}=18mol(NaHC{{O}_{3}})$
Calculate mol $N{{a}_{2}}C{{O}_{3}}$ by multiplying mol $NaHC{{O}_{3}}$ by the mol ratio between $N{{a}_{2}}C{{O}_{3}}$ and $NaHC{{O}_{3}}$ from the balanced equation, with $N{{a}_{2}}C{{O}_{3}}$ in the numerator.
$18mol(NaHC{{O}_{3}})\times \dfrac{1mol(NaHC{{O}_{3}})}{2mol(NaHC{{O}_{3}})}=9mol(NaHC{{O}_{3}})$
Calculate mass $N{{a}_{2}}C{{O}_{3}}$ by multiplying mol $N{{a}_{2}}C{{O}_{3}}$ by its molar mass $\left( 105.988\text{ }g/mol \right).$
$9mol(N{{a}_{2}}C{{O}_{3}})\times \dfrac{105.988g(N{{a}_{2}}C{{O}_{3}})}{1mol(N{{a}_{2}}C{{O}_{3}})}=950g(N{{a}_{2}}C{{O}_{3}})$
$1.5kg\left( NaHC{{O}_{3}} \right)$ will produce $\approx 950g\left( N{{a}_{2}}C{{O}_{3}} \right)$ upon decomposition.

Note:
Sodium bicarbonate is also known as baking soda whereas sodium carbonate is popularly known as washing soda or soda ash. Sodium bicarbonate is made up of sodium, acid and hydrogen whereas sodium carbonate is made up of sodium and acid. Although they sound similar with their names they are different.