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)

1 Answer
Oct 23, 2016

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.