Mg (50.0 g) and HCI (75.0 g) were added together to produce MgCl2 and H2 Mg + 2HCl → MgCl2 + H2
a) What mass of hydrogen gas produced in this reaction?
b) how much of the excess reactant is left over?

Mass of hydrogen produced  = 2.1 g

Mass of excess reactant left = 25.2 g

Explanation:

Given data:

Mass of Mg = 50.0 g

Mass of HCl = 75.0 g

Mass of hydrogen produced = ?

Mass of excess reactant left = ?

Solution:

Chemical equation:

Mg + 2HCl      →       MgCl₂ + H₂

Number of moles of Mg:

Number of moles = mass/molar mass

Number of moles = 50 g/ 24 g/mol

Number of moles = 2.1 mol

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 75 g/ 36.5 g/mol

Number of moles = 2.1 mol

now we will compare the moles of hydrogen gas with both reactant.

                     Mg       :        H₂

                       1          :       1

                     2.1         :     2.1

                   HCl         :        H₂

                      2          :          1

                    2.1          :        1/2×2.1 = 1.05 mol

HCl is limiting reactant and will limit the yield of hydrogen gas.

Mass of hydrogen:

Mass = number of moles × molar mass

Mass=  1.05 mol ×2 g/mol

Mass = 2.1 g

Mg is present in excess.

Mass of Mg left:

         HCl            :           Mg

             2             :            1

            2.1            :         1/2×2.1 = 1.05

Out of 2.1 moles of Mg 1.05 react with HCl.

Moles of Mg left = 2.1 mol - 1.05 mol = 1.05 mol

Mass of Mg left:

Mass = number of moles × molar mass

Mass = 1.05 mol × 24 g/mol

Mass = 25.2 g

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