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To solve the problem of how many grams of [tex]\( H_2O \)[/tex] can be formed when [tex]\( 6.12 \)[/tex] g of [tex]\( NH_3 \)[/tex] reacts with [tex]\( 3.78 \)[/tex] g of [tex]\( O_2 \)[/tex], follow these steps:
1. Calculate the molar masses:
- Molar mass of [tex]\( NH_3 \)[/tex] (Ammonia): [tex]\( 14.01 (\text{N}) + 3 \times 1.01 (\text{H}) = 17.03 \)[/tex] g/mol
- Molar mass of [tex]\( O_2 \)[/tex] (Oxygen): [tex]\( 2 \times 16.00 (\text{O}) = 32.00 \)[/tex] g/mol
- Molar mass of [tex]\( H_2O \)[/tex] (Water): [tex]\( 2 \times 1.01 (\text{H}) + 16.00 (\text{O}) = 18.02 \)[/tex] g/mol
2. Convert given masses to moles:
- Moles of [tex]\( NH_3 \)[/tex]: [tex]\( \frac{6.12 \, \text{g}}{17.03 \, \text{g/mol}} \approx 0.359 \, \text{mol} \)[/tex]
- Moles of [tex]\( O_2 \)[/tex]: [tex]\( \frac{3.78 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.118 \, \text{mol} \)[/tex]
3. Determine the stoichiometric ratios from the balanced equation:
[tex]\[ 4 \, NH_3 + 5 \, O_2 \rightarrow 4 \, NO + 6 \, H_2O \][/tex]
- According to the equation, 4 moles of [tex]\( NH_3 \)[/tex] produce 6 moles of [tex]\( H_2O \)[/tex].
- According to the equation, 5 moles of [tex]\( O_2 \)[/tex] produce 6 moles of [tex]\( H_2O \)[/tex].
4. Calculate the moles of [tex]\( H_2O \)[/tex] produced by each reactant:
- From [tex]\( NH_3 \)[/tex]: [tex]\( 0.359 \, \text{mol} \, NH_3 \times \frac{6 \, \text{mol} \, H_2O}{4 \, \text{mol} \, NH_3} \approx 0.539 \, \text{mol} \, H_2O \)[/tex]
- From [tex]\( O_2 \)[/tex]: [tex]\( 0.118 \, \text{mol} \, O_2 \times \frac{6 \, \text{mol} \, H_2O}{5 \, \text{mol} \, O_2} \approx 0.141 \, \text{mol} \, H_2O \)[/tex]
5. Identify the limiting reactant:
- The limiting reactant is [tex]\( O_2 \)[/tex] because it produces fewer moles of [tex]\( H_2O \)[/tex].
6. Calculate the mass of [tex]\( H_2O \)[/tex] formed:
- Moles of [tex]\( H_2O \)[/tex] from the limiting reactant: [tex]\( 0.141 \, \text{mol} \)[/tex]
- Mass of [tex]\( H_2O \)[/tex]: [tex]\( 0.141 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 2.55 \, \text{g} \)[/tex]
So, the correct answer is:
[tex]\[ \boxed{2.55 \, \text{g}} \][/tex]
1. Calculate the molar masses:
- Molar mass of [tex]\( NH_3 \)[/tex] (Ammonia): [tex]\( 14.01 (\text{N}) + 3 \times 1.01 (\text{H}) = 17.03 \)[/tex] g/mol
- Molar mass of [tex]\( O_2 \)[/tex] (Oxygen): [tex]\( 2 \times 16.00 (\text{O}) = 32.00 \)[/tex] g/mol
- Molar mass of [tex]\( H_2O \)[/tex] (Water): [tex]\( 2 \times 1.01 (\text{H}) + 16.00 (\text{O}) = 18.02 \)[/tex] g/mol
2. Convert given masses to moles:
- Moles of [tex]\( NH_3 \)[/tex]: [tex]\( \frac{6.12 \, \text{g}}{17.03 \, \text{g/mol}} \approx 0.359 \, \text{mol} \)[/tex]
- Moles of [tex]\( O_2 \)[/tex]: [tex]\( \frac{3.78 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.118 \, \text{mol} \)[/tex]
3. Determine the stoichiometric ratios from the balanced equation:
[tex]\[ 4 \, NH_3 + 5 \, O_2 \rightarrow 4 \, NO + 6 \, H_2O \][/tex]
- According to the equation, 4 moles of [tex]\( NH_3 \)[/tex] produce 6 moles of [tex]\( H_2O \)[/tex].
- According to the equation, 5 moles of [tex]\( O_2 \)[/tex] produce 6 moles of [tex]\( H_2O \)[/tex].
4. Calculate the moles of [tex]\( H_2O \)[/tex] produced by each reactant:
- From [tex]\( NH_3 \)[/tex]: [tex]\( 0.359 \, \text{mol} \, NH_3 \times \frac{6 \, \text{mol} \, H_2O}{4 \, \text{mol} \, NH_3} \approx 0.539 \, \text{mol} \, H_2O \)[/tex]
- From [tex]\( O_2 \)[/tex]: [tex]\( 0.118 \, \text{mol} \, O_2 \times \frac{6 \, \text{mol} \, H_2O}{5 \, \text{mol} \, O_2} \approx 0.141 \, \text{mol} \, H_2O \)[/tex]
5. Identify the limiting reactant:
- The limiting reactant is [tex]\( O_2 \)[/tex] because it produces fewer moles of [tex]\( H_2O \)[/tex].
6. Calculate the mass of [tex]\( H_2O \)[/tex] formed:
- Moles of [tex]\( H_2O \)[/tex] from the limiting reactant: [tex]\( 0.141 \, \text{mol} \)[/tex]
- Mass of [tex]\( H_2O \)[/tex]: [tex]\( 0.141 \, \text{mol} \times 18.02 \, \text{g/mol} \approx 2.55 \, \text{g} \)[/tex]
So, the correct answer is:
[tex]\[ \boxed{2.55 \, \text{g}} \][/tex]
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