Answer :
To solve this problem, we need to determine how much heat is produced when 38.1 g of methanol reacts with 59.1 g of oxygen according to the given thermochemical equation:
[tex]\[ CH_3OH(g) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \quad \Delta H = -764 \, \text{kJ} \][/tex]
This reaction tells us that burning 1 mole of methanol releases 764 kJ of heat. Here is the step-by-step solution:
1. Calculate the Molar Mass:
- Molar mass of methanol ([tex]\( CH_3OH \)[/tex]):
[tex]\[ 12 (\text{C}) + 3 \times 1 (\text{H}) + 16 (\text{O}) + 1 (\text{H}) = 32 \, \text{g/mol} \][/tex]
- Molar mass of oxygen ([tex]\( O_2 \)[/tex]):
[tex]\[ 2 \times 16 = 32 \, \text{g/mol} \][/tex]
2. Calculate the Moles of Each Reactant:
- Moles of methanol:
[tex]\[ \frac{38.1 \, \text{g}}{32 \, \text{g/mol}} \approx 1.19 \, \text{mol} \][/tex]
- Moles of oxygen:
[tex]\[ \frac{59.1 \, \text{g}}{32 \, \text{g/mol}} \approx 1.85 \, \text{mol} \][/tex]
3. Determine the Limiting Reactant:
- According to the balanced equation, 1 mole of methanol reacts with [tex]\( \frac{3}{2} \)[/tex] or 1.5 moles of oxygen.
- Determine how many moles of oxygen are needed for 1.19 moles of methanol:
[tex]\[ 1.19 \, \text{mol} \times 1.5 = 1.785 \, \text{mol of } O_2 \][/tex]
- Compare this with the available 1.85 moles of [tex]\( O_2 \)[/tex]. Since 1.85 moles of [tex]\( O_2 \)[/tex] is more than 1.785 moles needed, oxygen is present in excess.
- Therefore, methanol is the limiting reactant.
4. Calculate the Heat Produced:
- Since methanol is the limiting reactant and can fully react, the heat produced is based on the moles of methanol.
- The reaction enthalpy ([tex]\( \Delta H \)[/tex]) is [tex]\(-764 \, \text{kJ/mol}\)[/tex].
- The heat produced by 1.19 moles of methanol is:
[tex]\[ 1.19 \, \text{mol} \times (-764 \, \text{kJ/mol}) \approx -909.64 \, \text{kJ} \][/tex]
Based on these calculations, the heat produced is approximately [tex]\(-909.64 \, \text{kJ}\)[/tex], and methanol is the limiting reactant.
[tex]\[ CH_3OH(g) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \quad \Delta H = -764 \, \text{kJ} \][/tex]
This reaction tells us that burning 1 mole of methanol releases 764 kJ of heat. Here is the step-by-step solution:
1. Calculate the Molar Mass:
- Molar mass of methanol ([tex]\( CH_3OH \)[/tex]):
[tex]\[ 12 (\text{C}) + 3 \times 1 (\text{H}) + 16 (\text{O}) + 1 (\text{H}) = 32 \, \text{g/mol} \][/tex]
- Molar mass of oxygen ([tex]\( O_2 \)[/tex]):
[tex]\[ 2 \times 16 = 32 \, \text{g/mol} \][/tex]
2. Calculate the Moles of Each Reactant:
- Moles of methanol:
[tex]\[ \frac{38.1 \, \text{g}}{32 \, \text{g/mol}} \approx 1.19 \, \text{mol} \][/tex]
- Moles of oxygen:
[tex]\[ \frac{59.1 \, \text{g}}{32 \, \text{g/mol}} \approx 1.85 \, \text{mol} \][/tex]
3. Determine the Limiting Reactant:
- According to the balanced equation, 1 mole of methanol reacts with [tex]\( \frac{3}{2} \)[/tex] or 1.5 moles of oxygen.
- Determine how many moles of oxygen are needed for 1.19 moles of methanol:
[tex]\[ 1.19 \, \text{mol} \times 1.5 = 1.785 \, \text{mol of } O_2 \][/tex]
- Compare this with the available 1.85 moles of [tex]\( O_2 \)[/tex]. Since 1.85 moles of [tex]\( O_2 \)[/tex] is more than 1.785 moles needed, oxygen is present in excess.
- Therefore, methanol is the limiting reactant.
4. Calculate the Heat Produced:
- Since methanol is the limiting reactant and can fully react, the heat produced is based on the moles of methanol.
- The reaction enthalpy ([tex]\( \Delta H \)[/tex]) is [tex]\(-764 \, \text{kJ/mol}\)[/tex].
- The heat produced by 1.19 moles of methanol is:
[tex]\[ 1.19 \, \text{mol} \times (-764 \, \text{kJ/mol}) \approx -909.64 \, \text{kJ} \][/tex]
Based on these calculations, the heat produced is approximately [tex]\(-909.64 \, \text{kJ}\)[/tex], and methanol is the limiting reactant.
