To determine how many moles of hydrogen gas (H2) can be produced when 3.40 moles of hydrochloric acid (HCl) react with an excess of aluminum (Al), we need to use stoichiometry based on the balanced chemical equation:
[tex]\[ 6 \text{ HCl} + 2 \text{ Al} \rightarrow 3 \text{ H}_2 + 2 \text{ AlCl}_3 \][/tex]
1. Identify the mole ratio from the balanced equation:
From the balanced chemical equation, we see that 6 moles of HCl produce 3 moles of H2. The ratio of moles of HCl to moles of H2 is therefore 6:3, which simplifies to 2:1.
2. Given moles of HCl:
We are given that 3.40 moles of HCl react.
3. Use the mole ratio to find moles of H2:
Since the ratio of HCl to H2 is 2:1, we can use this ratio to determine the moles of H2 produced.
[tex]\[ \text{Moles of H}_2 = \left(\frac{\text{Moles of HCl}}{2}\right) \times 1 \][/tex]
4. Calculate:
Substitute 3.40 moles of HCl into the equation:
[tex]\[ \text{Moles of H}_2 = \left(\frac{3.40}{2}\right) \times 1 \][/tex]
[tex]\[ \text{Moles of H}_2 = 1.70 \][/tex]
Therefore, 1.70 moles of hydrogen gas (H2) can be produced when 3.40 moles of HCl react with an excess of aluminum (Al).
