2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams of H2 are also produced?
2 answers:
So,
Our conceptual plan is as follows:
g AlCl3 --> mol AlCl3 --> mol H2 --> g H2
Hope this helps!
Answer:
B. 2.92
Explanation:
It is a stichiometry problem. From the balanced given reaction: 2Al + 6HCl → 2AlCl₃ + 3H₂, It is clear that 2.o moles of Al reacts with 6.0 moles of HCl to produce 2.0 moles of AlCl₃ and 3.0 moles of H₂. Herein, we are concerned with the two products that the reaction results <em>2.0 moles of AlCl₃ with 2.0 moles of H₂.</em> So, we should convert the amount of grams of AlCl₃ (129.0 g) to number of moles (n) using the relation: <em>n = mass / molar mass,</em>
∴ n of AlCl₃ = (129.0 g) / (133.34 g/mol) = 0.967 mol.
<em><u>Using cross multiplication:</u></em>
2.0 moles of AlCl₃ produced with → 3.0 moles of H₂, from the stichiometry.
0.967 moles of AlCl₃ produced with → ??? moles of H₂.
∴ the number of moles of H₂ = (0.967)(3.0) / (2.0) = 1.45 mol.
Now, we can get the grams of H₂: ∴ The grams of H₂ = n x molar mass = (1.45 mol)(2.01588 g/mol) = 2.92 g.
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