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Answer :
Let's solve the problem step by step to find the mass of [tex]\( \text{CO}_2 \)[/tex] produced when 68.0 g of [tex]\( \text{CaO} \)[/tex] is reacted with 50.0 g of [tex]\( \text{C} \)[/tex] according to the reaction:
[tex]\[ \text{CaO} + \text{C} \rightarrow \text{CaC}_2 + \text{CO}_2 \][/tex]
### 1. Find the molar masses:
- Molar mass of [tex]\( \text{CaO} \)[/tex] is approximately 56.08 g/mol.
- Molar mass of [tex]\( \text{C} \)[/tex] is approximately 12.01 g/mol.
- Molar mass of [tex]\( \text{CO}_2 \)[/tex] is approximately 44.01 g/mol.
### 2. Calculate the moles of each reactant:
For [tex]\( \text{CaO} \)[/tex]:
[tex]\[
\text{Moles of CaO} = \frac{68.0 \, \text{g}}{56.08 \, \text{g/mol}} \approx 1.2125 \, \text{mol}
\][/tex]
For [tex]\( \text{C} \)[/tex]:
[tex]\[
\text{Moles of C} = \frac{50.0 \, \text{g}}{12.01 \, \text{g/mol}} \approx 4.1632 \, \text{mol}
\][/tex]
### 3. Determine the limiting reactant:
According to the balanced reaction:
[tex]\[ \text{CaO} + 3\text{C} \rightarrow \text{CaC}_2 + \text{CO}_2 \][/tex]
- 1 mole of [tex]\( \text{CaO} \)[/tex] reacts with 3 moles of [tex]\( \text{C} \)[/tex].
Since we have approximately 1.2125 moles of [tex]\( \text{CaO} \)[/tex], it would need:
[tex]\[ 1.2125 \text{ moles of CaO} \times 3 \text{ moles of C/mole of CaO} = 3.6375 \text{ moles of C} \][/tex]
We have 4.1632 moles of [tex]\( \text{C} \)[/tex], which is more than needed. Therefore, [tex]\( \text{CaO} \)[/tex] is the limiting reactant.
### 4. Calculate the moles of [tex]\( \text{CO}_2 \)[/tex] produced:
According to the stoichiometry of the reaction, 1 mole of [tex]\( \text{CaO} \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex]. Thus, the moles of [tex]\( \text{CO}_2 \)[/tex] produced will be equal to the moles of [tex]\( \text{CaO} \)[/tex].
[tex]\[
\text{Moles of CO}_2 = 1.2125 \, \text{mol}
\][/tex]
### 5. Calculate the mass of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[
\text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 = 1.2125 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 53.36 \, \text{g}
\][/tex]
Thus, approximately 53.36 g of [tex]\( \text{CO}_2 \)[/tex] is produced when 68.0 g of [tex]\( \text{CaO} \)[/tex] is reacted with 50.0 g of [tex]\( \text{C} \)[/tex].
[tex]\[ \text{CaO} + \text{C} \rightarrow \text{CaC}_2 + \text{CO}_2 \][/tex]
### 1. Find the molar masses:
- Molar mass of [tex]\( \text{CaO} \)[/tex] is approximately 56.08 g/mol.
- Molar mass of [tex]\( \text{C} \)[/tex] is approximately 12.01 g/mol.
- Molar mass of [tex]\( \text{CO}_2 \)[/tex] is approximately 44.01 g/mol.
### 2. Calculate the moles of each reactant:
For [tex]\( \text{CaO} \)[/tex]:
[tex]\[
\text{Moles of CaO} = \frac{68.0 \, \text{g}}{56.08 \, \text{g/mol}} \approx 1.2125 \, \text{mol}
\][/tex]
For [tex]\( \text{C} \)[/tex]:
[tex]\[
\text{Moles of C} = \frac{50.0 \, \text{g}}{12.01 \, \text{g/mol}} \approx 4.1632 \, \text{mol}
\][/tex]
### 3. Determine the limiting reactant:
According to the balanced reaction:
[tex]\[ \text{CaO} + 3\text{C} \rightarrow \text{CaC}_2 + \text{CO}_2 \][/tex]
- 1 mole of [tex]\( \text{CaO} \)[/tex] reacts with 3 moles of [tex]\( \text{C} \)[/tex].
Since we have approximately 1.2125 moles of [tex]\( \text{CaO} \)[/tex], it would need:
[tex]\[ 1.2125 \text{ moles of CaO} \times 3 \text{ moles of C/mole of CaO} = 3.6375 \text{ moles of C} \][/tex]
We have 4.1632 moles of [tex]\( \text{C} \)[/tex], which is more than needed. Therefore, [tex]\( \text{CaO} \)[/tex] is the limiting reactant.
### 4. Calculate the moles of [tex]\( \text{CO}_2 \)[/tex] produced:
According to the stoichiometry of the reaction, 1 mole of [tex]\( \text{CaO} \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex]. Thus, the moles of [tex]\( \text{CO}_2 \)[/tex] produced will be equal to the moles of [tex]\( \text{CaO} \)[/tex].
[tex]\[
\text{Moles of CO}_2 = 1.2125 \, \text{mol}
\][/tex]
### 5. Calculate the mass of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[
\text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2 = 1.2125 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 53.36 \, \text{g}
\][/tex]
Thus, approximately 53.36 g of [tex]\( \text{CO}_2 \)[/tex] is produced when 68.0 g of [tex]\( \text{CaO} \)[/tex] is reacted with 50.0 g of [tex]\( \text{C} \)[/tex].
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