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 liber

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

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Diễm Thu 5 years 2021-07-29T19:24:07+00:00 1 Answers 79 views 1

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  1. sori
    0
    2021-07-29T19:26:01+00:00
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    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.

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