QuestionSeptember 20, 2025

7. Acetylene (C_(2)H_(2)) is formed by reacting 7.08 g of C and 4.92 g of H_(2) What is the limiting reactant? How much of the other reactant is in excess? (hint: calculate how much reacts and subtract from the starting amount) 2C(s)+H_(2)(g)arrow C_(2)H_(2)(g)

7. Acetylene (C_(2)H_(2)) is formed by reacting 7.08 g of C and 4.92 g of H_(2) What is the limiting reactant? How much of the other reactant is in excess? (hint: calculate how much reacts and subtract from the starting amount) 2C(s)+H_(2)(g)arrow C_(2)H_(2)(g)
7. Acetylene (C_(2)H_(2)) is formed by reacting 7.08 g of C and 4.92 g of H_(2) What is the limiting reactant? How much of the other reactant is in excess? (hint: calculate how much reacts and subtract from the starting amount) 2C(s)+H_(2)(g)arrow C_(2)H_(2)(g)

Solution
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Answer

Limiting reactant: Carbon (C); Excess H_2 remaining: 4.32\,g Explanation 1. Calculate moles of each reactant n_C = \frac{7.08\,g}{12.01\,g/mol} = 0.590\,mol; n_{H_2} = \frac{4.92\,g}{2.016\,g/mol} = 2.44\,mol 2. Determine limiting reactant using stoichiometry Reaction: 2C + H_2 \rightarrow C_2H_2. For 0.590\,mol C, needed = 0.295\,mol H_2 (\frac{0.590}{2}). Available H_2 is 2.44\,mol, so C is limiting. 3. Calculate excess H_2 remaining 0.295\,mol H_2 reacts. Excess = 2.44 - 0.295 = 2.145\,mol. Mass = 2.145\,mol \times 2.016\,g/mol = 4.32\,g

Explanation

1. Calculate moles of each reactant<br /> $n_C = \frac{7.08\,g}{12.01\,g/mol} = 0.590\,mol$; $n_{H_2} = \frac{4.92\,g}{2.016\,g/mol} = 2.44\,mol$<br />2. Determine limiting reactant using stoichiometry<br /> Reaction: $2C + H_2 \rightarrow C_2H_2$. For $0.590\,mol$ C, needed $= 0.295\,mol$ $H_2$ ($\frac{0.590}{2}$). Available $H_2$ is $2.44\,mol$, so C is limiting.<br />3. Calculate excess $H_2$ remaining<br /> $0.295\,mol$ $H_2$ reacts. Excess $= 2.44 - 0.295 = 2.145\,mol$. Mass $= 2.145\,mol \times 2.016\,g/mol = 4.32\,g$
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