Question

Определить массу и объём водорода который образуется при взаимодействии 24 г магния с соляной кислотой

Answer 1

Mg+2HCl=MgCl2+H2
Дано:
m(Mg)=24 г
Vm=22.4 л/моль
Решение:
n(Mg)=m/M=24/24=1 моль
n(Н2)=1 моль (по уравнению хим. реакции)
m(Н2)=n*M=1*2=2 г
V(Н2)=n*Vm=1*22.4=22.4 л

Answer 2

Calculation of the Mass and Volume of Hydrogen Produced

To determine the mass and volume of hydrogen gas produced when 24 g of magnesium reacts with hydrochloric acid (HCl), we can use the balanced chemical equation for the reaction:

Mg + 2HCl → MgCl2 + H2

According to the equation, 1 mole of magnesium (Mg) reacts with 2 moles of hydrochloric acid (HCl) to produce 1 mole of magnesium chloride (MgCl2) and 1 mole of hydrogen gas (H2).

1. Calculate the moles of magnesium (Mg): - The molar mass of magnesium (Mg) is 24 g/mol. - Moles of magnesium (Mg) = Mass of magnesium (Mg) / Molar mass of magnesium (Mg) = 24 g / 24 g/mol = 1 mol.

2. Calculate the moles of hydrogen gas (H2): - According to the balanced equation, the ratio of moles of magnesium (Mg) to moles of hydrogen gas (H2) is 1:1. - Therefore, the moles of hydrogen gas (H2) produced is also 1 mol.

3. Calculate the mass of hydrogen gas (H2): - The molar mass of hydrogen gas (H2) is 2 g/mol. - Mass of hydrogen gas (H2) = Moles of hydrogen gas (H2) * Molar mass of hydrogen gas (H2) = 1 mol * 2 g/mol = 2 g.

4. Calculate the volume of hydrogen gas (H2): - The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L/mol. - Volume of hydrogen gas (H2) = Moles of hydrogen gas (H2) * Molar volume of hydrogen gas (H2) = 1 mol * 22.4 L/mol = 22.4 L.

Therefore, when 24 g of magnesium reacts with hydrochloric acid, it produces 2 g of hydrogen gas and occupies a volume of 22.4 L.

Please note that these calculations assume ideal conditions and may not account for any deviations from the ideal gas law or other factors that may affect the reaction.