Stoichiometry Moles equations

Question

Al + 3 NaOH → Al(OH)₃ + 3 Na

In the balanced chemical equation above, how many grams of Al(OH)₃ will be produced if 6.82 grams of NaOH reacts completely

Answer 1

Al + 3 NaOH → Al(OH)₃ + 3 Na

Assuming that the Al is not in limiting supply, we can find the moles of Al(OH)₃ produced from 6.82 g of NaOH, and then convert that to grams of Al(OH)₃.

moles Al(OH)₃ produced:

6.82 g NaOH x 1 mole NaOH/40.0 g = 0.1705 moles NaOH

From the balanced equation, 3 moles NaOH produces 1 moles Al(OH)₃. Thus...

0.1705 moles NaOH x 1 mole Al(OH)₃ / 3 moles NaOH = 0.0568 mols Al(OH)₃ produced

mass of Al(OH)₃ produced:

0.0568 moles Al(OH)₃ x 78 g /mole = 4.43 g Al(OH)₃ produced