In the following chemical reaction between H_2 and Cl_2 to produce HCl, what is the sum of the mass of HCl produced plus the mass of left over reactants when 0.40 g of H_2 completely reacts with 12.35 g of Cl_2? H_2(g) + Cl_2(g) → 2HCl(g)

Left over mass of hydrogen = 0.06 g

Mass of HCl produced = 12.41 g

Explanation:

Given data:

Mass of H₂ = 0.40 g

Mass of Cl₂ = 12.35 g

Mass of left over reactant = ?

Mass of HCl produced = ?

Solution:

Chemical equation:

H₂ + Cl₂    →     2HCl

Number of moles of H₂:

Number of moles = mass / molar mass

Number of moles = 0.40 g/ 2 g/mol

Number of moles = 0.2 mol

Number of moles of Cl₂:

Number of moles = mass / molar mass

Number of moles = 12.35 g/ 71 g/mol

Number of moles = 0.17 mol

Now we will compare the moles of HCl with H₂ and Cl₂.

                     H₂             :            HCl

                      1               :               2

                   0.2              :           2×0.2 = 0.4

                     Cl₂             :             HCl

                      1               :              2

                    0.17           :            2 × 0.17 = 0.34

Chlorine is limiting reactant.

Mass of HCl produced:

Mass = number of moles × molar mass

Mass = 0.34 mol × 36.5 g/mol

Mass = 12.41 g

Leftover mass of hydrogen:

                          Cl₂               :               H₂

                            1                :                 1

                           0.17            :             0.17

Number of moles of H₂ react with Cl₂  are 0.17.

Moles remain unreacted = 0.2 - 0.17 = 0.03 mol

Mass left over:

Mass = number of moles × molar mass

Mass = 0.03 mol × 2 g/mol

Mass = 0.06 g

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