How many moles of Al(OH)₃ are produced when 50.0 g of KOH react completely in the following equation? AlCl₃ + 3KOH $\rarr$ Al(OH)₃ + 3KCl

Question

Quizplus · Accepted Answer

First, convert 50.0 g of KOH to moles: $ \frac{50.0 \, \text{g}}{56.11 \, \text{g/mol}} = 0.891 \, \text{moles of KOH} $. According to the balanced chemical equation, 3 moles of KOH produce 1 mole of $ \text{Al(OH)}_3 $, so $ \frac{0.891 \, \text{moles of KOH}}{3} = 0.297 \, \text{moles of} \, \text{Al(OH)}_3 $.

How Many Moles of Al(OH)3 Are Produced When 50 →\rarr→ Al(OH)3 + 3KCl A)0

How Many Moles of Al(OH)₃ Are Produced When 50 $\rarr$ Al(OH)₃ + 3KCl
A)0