In the reaction $2 A l + 6 H B r \to 2 A l B r_{3} + 3 H_{2}$ $2Al +6HBr -> 2AlBr_3 + 3H_2$ , when 3.22 moles of $A l$ $Al$ reacts with 4.96 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?

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In the reaction $2 A l + 6 H B r \to 2 A l B r_{3} + 3 H_{2}$ $2Al +6HBr -> 2AlBr_3 + 3H_2$ , when 3.22 moles of $A l$ $Al$ reacts with 4.96 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?

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Answer 1 · 2016-02-05T21:40:18

$2.48 m o l$ $2.48mol$

Explanation:

The balanced chemical equation represents the mole ratio in which the chemicals react.

Since 2 moles aluminuim are required for every 6 moles of hydrogen bromide, yet we only have 3.22 moles aluminium and 4.96 moles hydrogen bromide, it implies that $H B r$ $HBr$ will get used up first and is hence the limiting reactant, while $A l$ $Al$ is in excess.

So all $4.96 m o l H B r$ $4.96 mol HBr$ is used up and reacts with $\frac{4.96}{3} = 1.653 m o l A l$ $4.96/3=1.653mol Al$ to produce $\frac{4.96}{2} = 2.48 m o l H_{2}$ $4.96/2=2.48 mol H_2$ .

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In the reaction $2 A l + 6 H B r \to 2 A l B r_{3} + 3 H_{2}$ $2Al +6HBr -> 2AlBr_3 + 3H_2$ , when 3.22 moles of $A l$ $Al$ reacts with 4.96 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?

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Explanation:

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In the reaction 2Al +6HBr -> 2AlBr_3 + 3H_22Al+6HBr→2AlBr3+3H22Al +6HBr -> 2AlBr_3 + 3H_2, when 3.22 moles of AlAlAl reacts with 4.96 moles of HBrHBrHBr, how many moles of H_2H2H_2 are formed?

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In the reaction $2 A l + 6 H B r \to 2 A l B r_{3} + 3 H_{2}$ $2Al +6HBr -> 2AlBr_3 + 3H_2$ , when 3.22 moles of $A l$ $Al$ reacts with 4.96 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?