How many liters of hydrogen gas are collected over water

Step 1 of 3

Here we have to find out the volume of \(\mathrm{H}_{2}\) gas collected and the mass of \(\mathrm{LiOH}\).

A balanced chemical equation for the reaction between \(\mathrm{Li}\) and water is given below;

                               \(2 \mathrm{Li}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{LiOH}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g})\)

Mass of \(\mathrm{Li}=0.84 \mathrm{~g}\)

Pressure \(=725 \mathrm{~mm} \mathrm{Hg}\)

\(1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg}\)

                               \(725 \mathrm{~mm} \mathrm{Hg}=725 \mathrm{~mm} \mathrm{Hg} \times \frac{1 \mathrm{~atm}}{760 \mathrm{mmHg}}=0.953 \mathrm{~atm}\)

Temperature \(=18^{\circ} \mathrm{C}=273+18=291 \mathrm{~K}\)

From the above balance equation it has been found that 1 mole of \(\mathrm{H}_{2}\) has formed from \(2 \mathrm{~mol}\) of Li.

Mole of \(\mathrm{H}_{2}\) can be calculated as,

                                 \(\text { Mass of } \mathrm{Li} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2}}{2 \mathrm{~mol} \mathrm{Li}}\)

                                        \(=0.84 \mathrm{~g} \text { of } \mathrm{Li} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2}}{2 \times 6.941 \mathrm{gLi}}\)

                                                           \(=0.06 \mathrm{~mol} \mathrm{H}_{2}\)