Firstly we must wright the reaction:
2HNO3 (dilute) + MgO = Mg(NO3)2 + H2O
Now we count out the amount of MgO
n = m/Mr = 2/40 = 0.05 mol
The amount of HNO3 :
n = 0.05 × 2 = 0.1 mol;
The volume of nitric acid is:
V = n/c = 0.1/2.5 = 0.04 dm3 = 0.04 l.
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