Answer :
To find the percent yield of ethane ([tex]$C_2H_6$[/tex]) when [tex]$31.0$[/tex] grams of acetylene ([tex]$C_2H_2$[/tex]) reacts with hydrogen, producing [tex]$22.7$[/tex] grams of ethane, you can follow these steps:
1. Calculate the moles of acetylene ([tex]$C_2H_2$[/tex]) used:
- The molar mass of acetylene ([tex]$C_2H_2$[/tex]) is [tex]$26.04$[/tex] g/mol.
- Using the mass of acetylene provided ([tex]$31.0$[/tex] grams):
[tex]\[ \text{Moles of } C_2H_2 = \frac{\text{Mass of } C_2H_2}{\text{Molar mass of } C_2H_2} = \frac{31.0 \text{ g}}{26.04 \text{ g/mol}} \approx 1.1904761904761905 \text{ mol} \][/tex]
2. Determine the moles of ethane ([tex]$C_2H_6$[/tex]) produced theoretically based on stoichiometry:
- The balanced chemical equation is:
[tex]\[ C_2H_2(g) + 2H_2(g) \rightarrow C_2H_6(g) \][/tex]
- From the equation, 1 mole of [tex]$C_2H_2$[/tex] produces 1 mole of [tex]$C_2H_6$[/tex].
- Therefore, the moles of ethane theoretically produced is also [tex]$1.1904761904761905$[/tex] moles.
3. Calculate the theoretical mass of ethane ([tex]$C_2H_6$[/tex]):
- The molar mass of ethane ([tex]$C_2H_6$[/tex]) is [tex]$30.07$[/tex] g/mol.
- Using the moles of ethane:
[tex]\[ \text{Theoretical mass of } C_2H_6 = \text{Moles of } C_2H_6 \times \text{Molar mass of } C_2H_6 = 1.1904761904761905 \text{ mol} \times 30.07 \text{ g/mol} \approx 35.79761904761905 \text{ g} \][/tex]
4. Calculate the percent yield:
- The actual mass of ethane produced is [tex]$22.7$[/tex] grams.
- The percent yield is calculated using the formula:
[tex]\[ \text{Percent yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100 = \left(\frac{22.7 \text{ g}}{35.79761904761905 \text{ g}}\right) \times 100 \approx 63.41203857665446\% \][/tex]
Therefore, the percent yield of [tex]$C_2H_6$[/tex] (ethane) is:
[tex]\[ \boxed{63.41 \%} \][/tex]
1. Calculate the moles of acetylene ([tex]$C_2H_2$[/tex]) used:
- The molar mass of acetylene ([tex]$C_2H_2$[/tex]) is [tex]$26.04$[/tex] g/mol.
- Using the mass of acetylene provided ([tex]$31.0$[/tex] grams):
[tex]\[ \text{Moles of } C_2H_2 = \frac{\text{Mass of } C_2H_2}{\text{Molar mass of } C_2H_2} = \frac{31.0 \text{ g}}{26.04 \text{ g/mol}} \approx 1.1904761904761905 \text{ mol} \][/tex]
2. Determine the moles of ethane ([tex]$C_2H_6$[/tex]) produced theoretically based on stoichiometry:
- The balanced chemical equation is:
[tex]\[ C_2H_2(g) + 2H_2(g) \rightarrow C_2H_6(g) \][/tex]
- From the equation, 1 mole of [tex]$C_2H_2$[/tex] produces 1 mole of [tex]$C_2H_6$[/tex].
- Therefore, the moles of ethane theoretically produced is also [tex]$1.1904761904761905$[/tex] moles.
3. Calculate the theoretical mass of ethane ([tex]$C_2H_6$[/tex]):
- The molar mass of ethane ([tex]$C_2H_6$[/tex]) is [tex]$30.07$[/tex] g/mol.
- Using the moles of ethane:
[tex]\[ \text{Theoretical mass of } C_2H_6 = \text{Moles of } C_2H_6 \times \text{Molar mass of } C_2H_6 = 1.1904761904761905 \text{ mol} \times 30.07 \text{ g/mol} \approx 35.79761904761905 \text{ g} \][/tex]
4. Calculate the percent yield:
- The actual mass of ethane produced is [tex]$22.7$[/tex] grams.
- The percent yield is calculated using the formula:
[tex]\[ \text{Percent yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100 = \left(\frac{22.7 \text{ g}}{35.79761904761905 \text{ g}}\right) \times 100 \approx 63.41203857665446\% \][/tex]
Therefore, the percent yield of [tex]$C_2H_6$[/tex] (ethane) is:
[tex]\[ \boxed{63.41 \%} \][/tex]
