Question

Сколько грамм магния получили при восстановлении водородом 32 г оксида магния только обизательно дано найти ришение

Answer 1

Ответ:

Объяснение:

Дано:

m(MgO)=32г

m(Mg)-?

MgO + H2 = MgO + H2↑

n(MgO)=m(MgO):M(MgO)=32:40=0.8моль

n(MgO)=n(Mg)

m(Mg)=n(Mg)*(Mg)=0.8*24=19.2г

Answer 2

Magnesium Obtained from the Reduction of Magnesium Oxide by Hydrogen

To find the amount of magnesium obtained when 32g of magnesium oxide is reduced by hydrogen, we can use the chemical equation for the reaction and stoichiometry to calculate the amount of magnesium produced.

The balanced chemical equation for the reaction is: MgO + H2 → Mg + H2O

According to the stoichiometry of the balanced equation, 1 mole of magnesium oxide reacts with 1 mole of hydrogen to produce 1 mole of magnesium. The molar mass of magnesium oxide (MgO) is 40.3 g/mol, and the molar mass of magnesium (Mg) is 24.3 g/mol.

Calculation

1. Calculate the moles of magnesium oxide: - Moles of MgO = Mass of MgO / Molar mass of MgO - Moles of MgO = 32g / 40.3 g/mol - Moles of MgO ≈ 0.793 moles

2. According to the stoichiometry of the balanced equation, the moles of magnesium produced will be the same as the moles of magnesium oxide consumed.

3. Calculate the mass of magnesium produced: - Mass of Mg = Moles of Mg × Molar mass of Mg - Mass of Mg = 0.793 moles × 24.3 g/mol - Mass of Mg ≈ 19.27g

Answer

When 32g of magnesium oxide is reduced by hydrogen, approximately 19.27 grams of magnesium are obtained