Question

El metal Mg reacciona con el Hcl para producir gas Hidrógeno Mg +2HCl---MgCl2 + H2. ¿Qué volumen en Litros de H2 en condiciones STP se liberan cuando reaccionan 8.25g de Mg? * 8.60L 6.60L 7.60L 16.60L

Answer 1

Answer: 7.60 L of hydrogen gas will be liberated.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    ......(1)

Given mass of Mg = 8.25 g

Molar mass of Mg = 24.305 g/mol

Plugging values in equation 1:

$\text{Moles of Mg}=\frac{8.25g}{24.305g/mol}=0.340mol$

The given chemical equation follows:

$Mg+2HCl\rightarrow MgCl_2+H_2$

By the stoichiometry of reaction:

If 1 mole of magnesium produces 1 mole of hydrogen gas

So, 0.340 moles of magnesium will be produce = $\frac{1}{1}\times 0.340=0.340mol$ of hydrogen gas

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.340 moles of hydrogen gas will occupy = $\frac{22.4L}{1mol}\times 0.340mol=7.60L$ of hydrogen gas

Hence, 7.60 L of hydrogen gas will be liberated.