QuestionMay 15, 2025

2Al+6HBrarrow 2AlBr_(3)+3H_(2) When 4.65 moles of Al reacts with 8.36 moles of HBr how many moles of H2 are formed? Moles H2 square Limiting Reactant square

2Al+6HBrarrow 2AlBr_(3)+3H_(2) When 4.65 moles of Al reacts with 8.36 moles of HBr how many moles of H2 are formed? Moles H2 square Limiting Reactant square
2Al+6HBrarrow 2AlBr_(3)+3H_(2) When 4.65 moles of Al reacts with 8.36 moles of HBr how many moles of H2 are formed? Moles H2 square Limiting Reactant square

Solution
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Answer

Moles H_2: 4.18 ### Limiting Reactant: HBr Explanation 1. Determine the limiting reactant The balanced equation shows that 2 moles of Al react with 6 moles of HBr. Calculate the required moles of HBr for 4.65 moles of Al: \frac{6}{2} \times 4.65 = 13.95 moles. Since only 8.36 moles of HBr are available, HBr is the limiting reactant. 2. Calculate moles of H_2 produced From the balanced equation, 6 moles of HBr produce 3 moles of H_2. Therefore, 8.36 moles of HBr will produce \frac{3}{6} \times 8.36 = 4.18 moles of H_2.

Explanation

1. Determine the limiting reactant<br /> The balanced equation shows that 2 moles of Al react with 6 moles of HBr. Calculate the required moles of HBr for 4.65 moles of Al: $\frac{6}{2} \times 4.65 = 13.95$ moles. Since only 8.36 moles of HBr are available, HBr is the limiting reactant.<br /><br />2. Calculate moles of $H_2$ produced<br /> From the balanced equation, 6 moles of HBr produce 3 moles of $H_2$. Therefore, 8.36 moles of HBr will produce $\frac{3}{6} \times 8.36 = 4.18$ moles of $H_2$.
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