Question

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.]

`2Na(s) + 2H_2O(l) -> 2NaOH(aq) + H_2(g)`

Answer 1

`2.4*dm^3` dihydrogen gas will evolve.

Explanation:

`Na(s) + H_2O(l)rarr NaOH(aq) + 1/2H_2(g)uarr`

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

You reacted `"0.2 mol"` sodium metal, and thus `"0.1 mol"` dihydrogen gas will evolve. Given your conditions, `0.1*cancel(mol)xx24*dm^-3*cancel(mol^-1)` dihydrogen will be evolved, i.e. `2.4*dm^3` `H_2` gas.

Note that `1*dm^3` `-=` `10^-3*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*L` of dihydrogen.