Answer :
Sure, let's solve this step-by-step!
The balanced combustion reaction for acetylene ([tex]$C_2H_2$[/tex]) with oxygen ([tex]$O_2$[/tex]) is as follows:
[tex]\[ 2 C_2H_2 + 5 O_2 \rightarrow 4 CO_2 + 2 H_2O \][/tex]
Here's the step-by-step process:
1. Determine the masses and molar masses:
- Mass of [tex]$C_2H_2$[/tex]: 58.0 g
- Mass of [tex]$O_2$[/tex]: 81.0 g
The molar masses are:
- Molar mass of [tex]$C_2H_2$[/tex]: 26.04 g/mol
- Molar mass of [tex]$O_2$[/tex]: 32.00 g/mol
2. Calculate the moles of each reactant:
- Moles of [tex]$C_2H_2$[/tex]:
[tex]\[ \frac{58.0 \text{ g}}{26.04 \text{ g/mol}} = 2.228 \text{ mol} \][/tex]
- Moles of [tex]$O_2$[/tex]:
[tex]\[ \frac{81.0 \text{ g}}{32.00 \text{ g/mol}} = 2.531 \text{ mol} \][/tex]
3. Use stoichiometric coefficients from the balanced equation:
- For [tex]$C_2H_2$[/tex]: 2 moles are required.
- For [tex]$O_2$[/tex]: 5 moles are required.
4. Determine which reactant is the limiting reactant:
Find the moles of [tex]$O_2$[/tex] required to completely react with the given moles of [tex]$C_2H_2$[/tex]:
[tex]\[ \text{Moles of } O_2 = 2.228 \text{ mol} \times \frac{5 \text{ mol } O_2}{2 \text{ mol } C_2H_2} = 5.57 \text{ mol } O_2 \][/tex]
Compare this with the available moles of [tex]$O_2$[/tex]:
- Available moles of [tex]$O_2$[/tex]: 2.531 mol
Since the required moles of [tex]$O_2$[/tex] (5.57 mol) are more than the available moles (2.531 mol), [tex]$O_2$[/tex] is the limiting reactant.
5. Calculate the yield of [tex]$CO_2$[/tex]:
Using the limiting reactant ([tex]$O_2$[/tex]):
[tex]\[ \text{Yield of } CO_2 = \frac{2.531 \text{ mol } O_2}{5} \times 4 \times 44.01 \text{ g/mol} \][/tex]
Simplify the expression:
[tex]\[ \text{Yield of } CO_2 = 2.0248 \text{ mol } \times 44.01 \text{ g/mol} \][/tex]
[tex]\[ \text{Yield of } CO_2 = 89.1 \text{ g} \][/tex]
Thus, the limiting reactant is [tex]$O_2$[/tex] and the yield of [tex]$CO_2$[/tex] is 89.1 g. Therefore, the correct answer is:
[tex]\[ O_2, 89.1 \, \text{g} \][/tex]
The balanced combustion reaction for acetylene ([tex]$C_2H_2$[/tex]) with oxygen ([tex]$O_2$[/tex]) is as follows:
[tex]\[ 2 C_2H_2 + 5 O_2 \rightarrow 4 CO_2 + 2 H_2O \][/tex]
Here's the step-by-step process:
1. Determine the masses and molar masses:
- Mass of [tex]$C_2H_2$[/tex]: 58.0 g
- Mass of [tex]$O_2$[/tex]: 81.0 g
The molar masses are:
- Molar mass of [tex]$C_2H_2$[/tex]: 26.04 g/mol
- Molar mass of [tex]$O_2$[/tex]: 32.00 g/mol
2. Calculate the moles of each reactant:
- Moles of [tex]$C_2H_2$[/tex]:
[tex]\[ \frac{58.0 \text{ g}}{26.04 \text{ g/mol}} = 2.228 \text{ mol} \][/tex]
- Moles of [tex]$O_2$[/tex]:
[tex]\[ \frac{81.0 \text{ g}}{32.00 \text{ g/mol}} = 2.531 \text{ mol} \][/tex]
3. Use stoichiometric coefficients from the balanced equation:
- For [tex]$C_2H_2$[/tex]: 2 moles are required.
- For [tex]$O_2$[/tex]: 5 moles are required.
4. Determine which reactant is the limiting reactant:
Find the moles of [tex]$O_2$[/tex] required to completely react with the given moles of [tex]$C_2H_2$[/tex]:
[tex]\[ \text{Moles of } O_2 = 2.228 \text{ mol} \times \frac{5 \text{ mol } O_2}{2 \text{ mol } C_2H_2} = 5.57 \text{ mol } O_2 \][/tex]
Compare this with the available moles of [tex]$O_2$[/tex]:
- Available moles of [tex]$O_2$[/tex]: 2.531 mol
Since the required moles of [tex]$O_2$[/tex] (5.57 mol) are more than the available moles (2.531 mol), [tex]$O_2$[/tex] is the limiting reactant.
5. Calculate the yield of [tex]$CO_2$[/tex]:
Using the limiting reactant ([tex]$O_2$[/tex]):
[tex]\[ \text{Yield of } CO_2 = \frac{2.531 \text{ mol } O_2}{5} \times 4 \times 44.01 \text{ g/mol} \][/tex]
Simplify the expression:
[tex]\[ \text{Yield of } CO_2 = 2.0248 \text{ mol } \times 44.01 \text{ g/mol} \][/tex]
[tex]\[ \text{Yield of } CO_2 = 89.1 \text{ g} \][/tex]
Thus, the limiting reactant is [tex]$O_2$[/tex] and the yield of [tex]$CO_2$[/tex] is 89.1 g. Therefore, the correct answer is:
[tex]\[ O_2, 89.1 \, \text{g} \][/tex]
