2Al2O3 (s) + 3C (s) LaTeX: \longrightarrow ⟶ 4Al (s) + 3CO2 (g) If 821 g of Al2O3 reacts with an excess of carbon to produce 349 g of aluminum, what is the % yield of the reaction?

Percent yield = 79.79 %

Explanation:

Given data:

Mass of Al₂O₃ = 821 g

Mass of Al = 349 g

Percent yield = ?

Solution:

Chemical equation:

2Al₂O₃ + 3C     →      4Al  + 3CO₂

Number of moles of Al₂O₃:

Number of moles = mass/molar mass

Number of moles = 821 g/ 101.96 g/mol

Number of moles = 8.1 mol

Now we will compare the moles of Al with Al₂O₃.

                Al₂O₃        :         Al

                    2           :          4

                  8.1           :      4/2×8.1 = 16.2 mol

Mass of Al:

Mass = number of moles × molar mass

Mass = 16.2 mol ×  27 g/mol

Mass = 437.4 g

Percent yield:

Percent yield = (actual yield / theoretical yield)× 100

Percent yield =  (349 g/ 437.4 g) × 100

Percent yield = 79.79 %

answer; 4 electrons;