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
anor277
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.

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