Question

Use this balanced equation to help you solve:

Answer 1

1. 842g of NaOH will form 547.3 g of Al(OH)₃

2. The yield is 93.55%

Explanation:

3NaOH + Al → Al(OH)₃ + 3Na

1.

Molar mass of NaOH = 40 g/mol

Molar mass of Al = 27 g/mol

Molar mass of Al(OH)₃ = 78 g/mol

According to the balanced equation:

3 moles of NaOH requires 1 mole of Al to form 1 mole of Al(OH)₃

The ratio of NaOH : Al : Al(OH)₃ = 3 : 1 : 1

3 X 40 g of NaOH reacts with 27 g of Al to form 78 g of Al(OH)₃

120 g of NaOH + 27g of Al → 78 g of Al(OH)₃

120g of NaOH form 78g of Al(OH)₃

1g of NaOH will form $\frac{78}{120}$ g of Al(OH)₃

842g of NaOH will form $\frac{78}{120} X 842 g$ of Al(OH)₃

                                   = 547.3 g of Al(OH)₃

Therefore, 842g of NaOH will form 547.3 g of Al(OH)₃

2. Only 512 g of Al(OH)₃ is formed

Yield % = ?

$Yield = \frac{512}{547.3} X 100\\\\Yield = 93.55$

Therefore, the yield is 93.55%