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16 September, 08:38

In the following reaction: Mg + 2HCl → MgCl2 + H2 How many liters of H2 would be produced if you started with 24.3 g of Mg?

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  1. 16 September, 11:14
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    Equation is as follow,

    Mg + 2 HCl → MgCl ₂ + H₂

    According to equation 24.3 g of Mg when reacted with excess of HCl produces 22.4 L (1 mole) of H ₂ gas, this statement can be written as,

    24.3 g of Mg = 22.4 L of H₂

    So,

    If 24.3 g of Mg is reacted then X L of H₂ will produce,

    Again

    24.3 g of Mg = X L of H₂

    Solving for X,

    X = (24.3 g * 22.4 L) : 24.3 g

    X = 22.4 L of H₂

    Interesting! It means we are provided with exactly one mole of Mg, so obviously one mole of H₂ gas is produced which occupies 22.4 L of volume.
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