Question

3. When 312.7 g of Al₂(SO4)3 react with plenty of magnesium, what is the percent yield when 38.5 g of aluminum are

formed?

Answer 1

the percent yield is 78.2%

Step-by-step explanation:

The balanced equation for the reaction is:

2 Al₂(SO4)3 + 3 Mg → 3 MgSO4 + 4 Al

The molar mass of Al₂(SO4)3 is 342.15 g/mol.

First, calculate the number of moles of Al₂(SO4)3 present:

312.7 g / 342.15 g/mol = 0.913 mol Al₂(SO4)3

According to the balanced equation, 2 moles of Al₂(SO4)3 produce 4 moles of Al:

0.913 mol Al₂(SO4)3 x (4 mol Al / 2 mol Al₂(SO4)3) = 1.826 mol Al

The theoretical yield of Al is therefore:

1.826 mol Al x 26.98 g/mol = 49.24 g Al

The percent yield is:

(actual yield / theoretical yield) x 100%

Substituting the given values:

(38.5 g Al / 49.24 g Al) x 100% = 78.2%

Therefore, the percent yield is 78.2%.