2. Suppose 13.7 g of C2H2 reacts with 18.5 g O2 according to the reaction below. C2H2(g) + O2(g) → CO2(g) + H2O(ℓ) a. What is the mass of CO2 produced? b. What is the limiting reagent?

Mass of CO₂ produced = 20.328 g

Oxygen is limiting reagent.

Explanation:

Given data:

Mass of C₂H₂ = 13.7 g

Mass of O₂ = 18.5 g

Mass of CO₂ produced = ?

What is limiting reagent = ?

Solution:

Chemical equation:

2C₂H₂ + 5O₂      →     4CO₂ + 2H₂O

Number of moles of C₂H₂:

Number of moles = mass /molar mass

Number of moles = 13.7 g/ 26.04 g/mol

Number of moles = 0.526 mol

Number of moles of O₂:

Number of moles = mass /molar mass

Number of moles = 18.5 g/ 32 g/mol

Number of moles = 0.578 mol

Now we will compare the moles of CO₂ with C₂H₂ and O₂

                      C₂H₂              :                CO₂

                           2                :                 4

                       0.526            :              4/2×0.526 = 1.052

                      O₂                  :                   CO₂

                        5                   :                     4

                      0.578             :                4/5×0.578 = 0.462

The number of moles of  CO₂ produced by O₂  are less thus oxygen will be limiting reactant.

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 0.462 mol × 44 g/mol

Mass = 20.328 g

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