Question

Calculate the volume of carbon dioxide obtained at NTP by heating 8.4 g of sodium bicarbonate.

Answer 1

$\begin{array}{c}{2NaHC{O}_3}_{\left(8.4g\right)}N{a}_{2}C{O}_3+H_{2}O+C{o}_2\uparrow \\ Here,\\ 2moleofNaHC{O}_{3}gives1moleofC{O}_{2}gas\\ so,\left(\frac{8.4}{84}\right)molesofNaHC{O}_{3}gives=\frac{1}{2}\times \left(\frac{8.4}{84}\right)moles\\ Since,1moles\left(gives\right)=22.4L\\ \therefore \frac{1}{2}\times \left(\frac{8.4}{84}\right)g=\frac{1}{2}\times \left(\frac{8.4}{84}\right)\times 22.4L\\ =1.12L\end{array}$

Which is the required answer.