What is the volume of hydrogen produced at room temperature and pressure, when 0.2mol of sodium is reacted with an excess of water? [1 mol of gas occupies 24 dm^3 at room temperature and pressure.]

$2 N a \left(s\right) + 2 {H}_{2} O \left(l\right) \to 2 N a O H \left(a q\right) + {H}_{2} \left(g\right)$

Oct 23, 2016

$2.4 \cdot {\mathrm{dm}}^{3}$ dihydrogen gas will evolve.

Explanation:

$N a \left(s\right) + {H}_{2} O \left(l\right) \rightarrow N a O H \left(a q\right) + \frac{1}{2} {H}_{2} \left(g\right) \uparrow$

You equation clearly shows that each equiv sodium metal produces 1/2 an equiv of dihydrogen gas.

You reacted $\text{0.2 mol}$ sodium metal, and thus $\text{0.1 mol}$ dihydrogen gas will evolve. Given your conditions, $0.1 \cdot \cancel{m o l} \times 24 \cdot {\mathrm{dm}}^{-} 3 \cdot \cancel{m o {l}^{-} 1}$ dihydrogen will be evolved, i.e. $2.4 \cdot {\mathrm{dm}}^{3}$ ${H}_{2}$ gas.

Note that $1 \cdot {\mathrm{dm}}^{3}$ $\equiv$ ${10}^{-} 3 \cdot {m}^{3}$ (because 1*dm^3=(10^-1*m)^3=10^-3*m^3=1*L);"there are 1000 litres in a cubic metre." And thus $2.4 \cdot L$ of dihydrogen.