When 4.22 moles of $A l$ $Al$ reacts with 5.0 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?

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When 4.22 moles of $A l$ $Al$ reacts with 5.0 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?

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Answer 1 · 2016-03-26T19:34:15

$2.5 m o l$ $2.5mol$

Explanation:

First write a balanced chemical equation for the reaction of the metal with the acid :

$2 A l + 6 H B r \to 2 A l B r_{3} + 3 H_{2}$ $2Al+6HBr->2AlBr_3+3H_2$ .

This represents the mole ratio in which the reactants combine and in which the products are formed.

Since $2$ $2$ moles $A l$ $Al$ requires $6$ $6$ moles $H B r$ $HBr$ , it implies that, given $4.22$ $4.22$ moles $A l$ $Al$ and $5.0$ $5.0$ moles $H B r$ $HBr$ , then $H B r$ $HBr$ is in excess and $A l$ $Al$ is the limiting reagent.

Hence all $5.0 m o l H B r$ $5.0 mol HBr$ will react with $\frac{5.0}{3} = 1.667 m o l A l$ $5.0/3=1.667mol Al$ to form $1.667 m o l A l B r_{3}$ $1.667mol AlBr_3$ and $\frac{5.0}{2} = 2.5 m o l H_{2}$ $5.0/2=2.5 mol H_2$

Answer 2 · 2016-03-26T19:36:22

$2.5$ $2.5$ $m o l$ $mol$ dihydrogen gas are formed.

Explanation:

$A l (s) + 3 H B r (a q) \to A l B r_{3} (a q) + \frac{3}{2} H_{2} (g) ↑ ⏐ ⏐ ⏐$ $Al(s) + 3HBr(aq) rarr AlBr_3(aq) + 3/2H_2(g)uarr$

Clearly, hydrobromic acid is in deficiency (complete rxn requires $> 12 \cdot m o l$ $>12*mol$ $H B r$ $HBr$ ). From the equation, $\frac{1}{2}$ $1/2$ $m o l$ $mol$ dihydrogen gas are evolved per mole of $H B r$ $HBr$ . Since $5$ $5$ $m o l$ $mol$ $H B r$ $HBr$ are used, half of this quantity of $H_{2}$ $H_2$ evolve.

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When 4.22 moles of $A l$ $Al$ reacts with 5.0 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?

2 Answers

Explanation:

Explanation:

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When 4.22 moles of AlAl reacts with 5.0 moles of HBrHBr, how many moles of H2H_2 are formed?

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Explanation:

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When 4.22 moles of $A l$ $Al$ reacts with 5.0 moles of $H B r$ $HBr$ , how many moles of $H_{2}$ $H_2$ are formed?