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Answer :
Certainly! Let's go through each question step by step to understand how the answers were determined:
### Problem 56:
We are asked how many moles of [tex]\( \text{H}_2\text{O} \)[/tex] are required to produce 4.5 moles of [tex]\( \text{NO}_3 \)[/tex]. The exact stoichiometry of the reaction wasn't provided, so without additional information, we assume a 1:1 ratio. Therefore, 4.5 moles of [tex]\( \text{H}_2\text{O} \)[/tex] would be needed to produce 4.5 moles of [tex]\( \text{NO}_3 \)[/tex].
### Problem 57:
This problem involves determining the limiting reactant when 1.6 grams of [tex]\( \text{CaC}_2 \)[/tex] and 1.8 grams of [tex]\( \text{H}_2\text{O} \)[/tex] are mixed and reacted.
- Molar Mass Calculation:
- [tex]\( \text{CaC}_2 \)[/tex]: Approximately 64.1 g/mol
- [tex]\( \text{H}_2\text{O} \)[/tex]: Approximately 18.0 g/mol
- Conversion to Moles:
- [tex]\( \text{Moles of CaC}_2 = \frac{1.6 \, \text{g}}{64.1 \, \text{g/mol}} = 0.02496 \, \text{moles} \)[/tex]
- [tex]\( \text{Moles of H}_2\text{O} = \frac{1.8 \, \text{g}}{18.0 \, \text{g/mol}} = 0.1 \, \text{moles} \)[/tex]
The balanced chemical equation suggests that each mole of [tex]\( \text{CaC}_2 \)[/tex] reacts with 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore:
- 0.02496 moles of [tex]\( \text{CaC}_2 \)[/tex] would need 0.04992 moles of [tex]\( \text{H}_2\text{O} \)[/tex], but we have 0.1 moles available. So, [tex]\( \text{CaC}_2 \)[/tex] is the limiting reactant.
The number of moles of the limiting reactant, [tex]\( \text{CaC}_2 \)[/tex], is approximately 0.02496 moles.
### Problem 58:
A student obtained 80 grams of [tex]\( \text{Cu}_2\text{S} \)[/tex] from heating 99.8 grams of copper in excess sulfur.
- Molar Mass Calculation:
- [tex]\( \text{Cu} \)[/tex]: 64 g/mol
- [tex]\( \text{Cu}_2\text{S} \)[/tex]: About 159.5 g/mol
- Theoretical Yield Calculation:
- Moles of Cu used: [tex]\( \frac{99.8 \, \text{g}}{64 \, \text{g/mol}} = 1.56 \, \text{moles} \)[/tex]
- The reaction stoichiometry typically requires 2 moles of Cu for 1 mole of [tex]\( \text{Cu}_2\text{S} \)[/tex], so:
- Theoretical moles of [tex]\( \text{Cu}_2\text{S} = 0.78 \, \text{moles} \)[/tex]
- Theoretical mass of [tex]\( \text{Cu}_2\text{S} = 0.78 \times 159.5 = 124.41 \, \text{grams} \)[/tex]
- Percentage Yield:
- Percentage yield = [tex]\( \frac{80 \, \text{g}}{124.41 \, \text{g}} \times 100 \approx 64.33\% \)[/tex]
### Problem 59:
For 171 grams of sucrose ([tex]\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)[/tex]):
- Molar Mass Calculation:
- Sucrose: Approximately 342.3 g/mol
- Moles Calculation:
- Moles of sucrose = [tex]\( \frac{171 \, \text{g}}{342.3 \, \text{g/mol}} \approx 0.5 \, \text{moles} \)[/tex]
### Problem 60:
For 1.5 moles of [tex]\( \text{CaCO}_3 \)[/tex]:
- Molar Mass Calculation:
- [tex]\( \text{CaCO}_3 \)[/tex]: Approximately 100.1 g/mol
- Mass Calculation:
- Mass of [tex]\( \text{CaCO}_3 = 1.5 \, \text{moles} \times 100.1 \, \text{g/mol} = 150.15 \, \text{grams} \)[/tex]
These steps outline the process used to solve each of the problems provided. If you have any further questions or need clarification, feel free to ask!
### Problem 56:
We are asked how many moles of [tex]\( \text{H}_2\text{O} \)[/tex] are required to produce 4.5 moles of [tex]\( \text{NO}_3 \)[/tex]. The exact stoichiometry of the reaction wasn't provided, so without additional information, we assume a 1:1 ratio. Therefore, 4.5 moles of [tex]\( \text{H}_2\text{O} \)[/tex] would be needed to produce 4.5 moles of [tex]\( \text{NO}_3 \)[/tex].
### Problem 57:
This problem involves determining the limiting reactant when 1.6 grams of [tex]\( \text{CaC}_2 \)[/tex] and 1.8 grams of [tex]\( \text{H}_2\text{O} \)[/tex] are mixed and reacted.
- Molar Mass Calculation:
- [tex]\( \text{CaC}_2 \)[/tex]: Approximately 64.1 g/mol
- [tex]\( \text{H}_2\text{O} \)[/tex]: Approximately 18.0 g/mol
- Conversion to Moles:
- [tex]\( \text{Moles of CaC}_2 = \frac{1.6 \, \text{g}}{64.1 \, \text{g/mol}} = 0.02496 \, \text{moles} \)[/tex]
- [tex]\( \text{Moles of H}_2\text{O} = \frac{1.8 \, \text{g}}{18.0 \, \text{g/mol}} = 0.1 \, \text{moles} \)[/tex]
The balanced chemical equation suggests that each mole of [tex]\( \text{CaC}_2 \)[/tex] reacts with 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore:
- 0.02496 moles of [tex]\( \text{CaC}_2 \)[/tex] would need 0.04992 moles of [tex]\( \text{H}_2\text{O} \)[/tex], but we have 0.1 moles available. So, [tex]\( \text{CaC}_2 \)[/tex] is the limiting reactant.
The number of moles of the limiting reactant, [tex]\( \text{CaC}_2 \)[/tex], is approximately 0.02496 moles.
### Problem 58:
A student obtained 80 grams of [tex]\( \text{Cu}_2\text{S} \)[/tex] from heating 99.8 grams of copper in excess sulfur.
- Molar Mass Calculation:
- [tex]\( \text{Cu} \)[/tex]: 64 g/mol
- [tex]\( \text{Cu}_2\text{S} \)[/tex]: About 159.5 g/mol
- Theoretical Yield Calculation:
- Moles of Cu used: [tex]\( \frac{99.8 \, \text{g}}{64 \, \text{g/mol}} = 1.56 \, \text{moles} \)[/tex]
- The reaction stoichiometry typically requires 2 moles of Cu for 1 mole of [tex]\( \text{Cu}_2\text{S} \)[/tex], so:
- Theoretical moles of [tex]\( \text{Cu}_2\text{S} = 0.78 \, \text{moles} \)[/tex]
- Theoretical mass of [tex]\( \text{Cu}_2\text{S} = 0.78 \times 159.5 = 124.41 \, \text{grams} \)[/tex]
- Percentage Yield:
- Percentage yield = [tex]\( \frac{80 \, \text{g}}{124.41 \, \text{g}} \times 100 \approx 64.33\% \)[/tex]
### Problem 59:
For 171 grams of sucrose ([tex]\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)[/tex]):
- Molar Mass Calculation:
- Sucrose: Approximately 342.3 g/mol
- Moles Calculation:
- Moles of sucrose = [tex]\( \frac{171 \, \text{g}}{342.3 \, \text{g/mol}} \approx 0.5 \, \text{moles} \)[/tex]
### Problem 60:
For 1.5 moles of [tex]\( \text{CaCO}_3 \)[/tex]:
- Molar Mass Calculation:
- [tex]\( \text{CaCO}_3 \)[/tex]: Approximately 100.1 g/mol
- Mass Calculation:
- Mass of [tex]\( \text{CaCO}_3 = 1.5 \, \text{moles} \times 100.1 \, \text{g/mol} = 150.15 \, \text{grams} \)[/tex]
These steps outline the process used to solve each of the problems provided. If you have any further questions or need clarification, feel free to ask!
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