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.]

Md Ariful Islam

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Md Ariful Islam

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

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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