Use this balanced equation to help you solve:
1. If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically form?
2. If only 512 grams actually form, what was the % yield? (Briefly describe the steps you would take to solve).

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 g of Al(OH)₃

842g of NaOH will form 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 % = ?

Therefore, the yield is 93.55%

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2