Sagot :
SOLUTION:
Step 1: Write the balanced chemical equation.
2Al + 6HBr → 3H₂ + 2AlBr₃
Step 2: Calculate the number of moles of H₂ formed by each reactant.
• Using Al
Based on the balanced chemical equation, 2 moles of Al is stoichiometrically equivalent to 3 mole of H₂.
[tex]\begin{aligned} \text{moles of} \: \text{H}_2 & = \text{3.22 mol Al} \times \frac{\text{3 mol} \: \text{H}_2}{\text{2 mol Al}} \\ & = \text{4.83 mol} \end{aligned}[/tex]
• Using HBr
Based on the balanced chemical equation, 6 moles of Hbr is stoichiometrically equivalent to 3 mole of H₂.
[tex]\begin{aligned} \text{moles of} \: \text{H}_2 & = \text{4.96 mol HBr} \times \frac{\text{3 mol} \: \text{H}_2}{\text{6 mol HBr}} \\ & = \text{2.48 mol} \end{aligned}[/tex]
Step 3: Determine the limiting reagent.
Since HBr produced less amount of H₂ than Al, HBr is the limiting reagent.
Step 4: Determine the mass of H₂ formed.
Note that the (maximum) mass of a product is dictated by the limiting reagent. In this case, we will start at the number of moles of H₂ formed from the limiting reagent (HBr) which is equal to 2.48 mol.
The molar mass of H₂ is 2.016 g.
[tex]\begin{aligned} \text{mass of} \: \text{H}_2 & = \text{2.48 mol} \: \text{H}_2 \times \frac{\text{2.016 g} \: \text{H}_2}{\text{1 mol} \: \text{H}_2} \\ & = \boxed{\text{5.00 g}} \end{aligned}[/tex]
Hence, 5.00 g of H₂ are formed.
[tex]\\[/tex]
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