2 KHCO_3 +CO_2+H_2 O find amount of gases formed (in lit). When...

2 \mathrm{KHCO}_3 \longrightarrow \ldots+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}

find amount of gases formed (in lit). When amount of

\mathrm{KHCO}_3

is 33 gm .

Options

C. 7.46

The given reaction is as follows

2 \mathrm{KHCO}_3 \longrightarrow \mathrm{K}_2 \mathrm{CO}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}

(l)

\because 2

moles of

\mathrm{KHCO}_3

gives

=1 \mathrm{~mol} of \mathrm{K}_2 \mathrm{CO}_3 \therefore 1

moles of

\mathrm{KHCO}_3

gives

=1 / 2 \mathrm{~mol} of \mathrm{K}_2 \mathrm{CO}_3 \therefore 33 \mathrm{gm}\left(\frac{33}{100} \mathrm{~mol}\right) of \mathrm{KHCO}_3

gives

=\frac{33}{200}

mol of

\mathrm{K}_2 \mathrm{CO}_3

Similarly,

33 \mathrm{gm}\left(\frac{33}{100} \mathrm{~mol}\right)

\mathrm{KHCO}_3

gives

=\frac{33}{200} \mathrm{~mol}

\mathrm{CO}_2

Total moles of gases

=\frac{33}{200}+\frac{33}{200}=\frac{66}{200} \mathrm{~mol}

\begin{aligned} 1 \mathrm{~mol} \text{\text{ of } gas } & =224 \mathrm{l} \text{ volume } \\ \therefore \quad \frac{66}{200} \mathrm{~mol} \text{ of gas } & =\frac{66}{200} \times 22.4=7.46 \mathrm{l} .\end{aligned}