Question

Какой объём водорода выделится в результате взаимодействия 72г магния с серной кислотой?

Answer 1

 Mg + H2SO4 (разб.) = MgSO4+H2
n(Mg)=m/M=72/24=3(моль)
над магнием ставишь 3 моль, под ним 1 моль
над водородом ставишь х моль, под ним 1 моль
3:1=х:1 => х=3 => n(H2)= 3 моль
V(H2)=n*Vm= 3*22,4=67,2(л)
Ответ: 67,2л

Answer 2

Reaction and Calculation

When 72g of magnesium reacts with sulfuric acid, the amount of hydrogen gas produced can be calculated using the stoichiometry of the reaction. The balanced chemical equation for the reaction is:

Mg + H2SO4 → MgSO4 + H2

According to the stoichiometry of the reaction, 1 mole of magnesium reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas. The molar mass of magnesium is 24.305 g/mol, and the molar mass of sulfuric acid is 98.079 g/mol.

Using this information, we can calculate the amount of hydrogen gas produced.

Calculation

First, we need to convert the mass of magnesium to moles: - Moles of Mg = Mass of Mg / Molar mass of Mg - Moles of Mg = 72g / 24.305 g/mol - Moles of Mg ≈ 2.96 moles

Since the reaction is 1:1 between magnesium and hydrogen gas, the moles of hydrogen gas produced will be the same as the moles of magnesium.

Result

Therefore, approximately 2.96 moles of hydrogen gas will be produced as a result of the reaction between 72g of magnesium and sulfuric acid.