Question

When $4$ g of a mixture of $NaHC{O}_3$ and $NaCl$ is heated, $0.66$ g $C{O}_2$ gas is evolved. The percentage composition of the original mixture by mass is;

• A. $NaHC{O}_3=63$%, $NaCl=37$%
• B. $NaHC{O}_3=42$%, $NaCl=58$%
• C. $NaHC{O}_3=31.5$%, $NaCl=68.5$%
• D. $NaHC{O}_3=37$%, $NaCl=63$%

Answer 1

The correct option is A $NaHC{O}_3=63$%, $NaCl=37$%

$2NaHC{O}_3=N{a}_{2}C{O}_3+C{O}_2+H_{2}O$

Molar mass of $NaHC{O}_3=23+1+12+16\times 3=84g/mol$

2 moles of $NaHC{O}_3=2\times 84=168g$

Molar mass of $C{O}_2=12+16\times 2=44g/mol$

44g of $C{O}_2$ is produced from 168g of $NaHC{O}_3$

0.66g of $C{O}_2$ will be produced from $=(0.66\times 168)/44$

$=2.52g$

Percentage of $NaHC{O}_3$ in the mixture $=\left(2.52/4\right)\times 100=63$%

Percentage of $NaCl$ in the mixture $=100–63=37$%