How much hydrogen will be released when 120 g of 15% sulfuric acid interacts with the amount

| August 10, 2021 | education

How much hydrogen will be released when 120 g of 15% sulfuric acid interacts with the amount of magnesium required for the reaction?

Let’s write the reaction equation:
Mg + H2SO4 = MgSO4 + H2
Let us find the mass of sulfuric acid m (H2SO4) = w * m (solution), where w is the mass fraction of sulfuric acid in the solution, m (p-pa) is the mass of the sulfuric acid solution.
m (H2SO4) = 0.15 * 120 = 18 g
Let us find the amount of sulfuric acid substance n (H2SO4) = m (H2SO4) / M (H2SO4), where M (H2SO4) = 98 g / mol is the molar mass of sulfuric acid.
n (H2SO4) = 18/98 = 0.184 mol
The reaction equation shows that the amount of hydrogen substance is n (H2) = n (H2SO4) = 0.184 mol.
Under normal conditions, the volume of hydrogen V (H2) = Vm * n (H2) = 22.4 * 0.184 = 4.1 l
Vm = 22.4 l / mol – molar volume



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