To determine the mass of hydrogen used in the reaction, we first need to find the molar masses of nitrogen (N₂), hydrogen (H₂), and ammonia (NH₃):
- The molar mass of N₂ (nitrogen) is approximately \( 28.0 , \text{g/mol} \).
- The molar mass of H₂ (hydrogen) is approximately \( 2.0 , \text{g/mol} \).
- The molar mass of NH₃ (ammonia) is approximately \( 17.0 , \text{g/mol} \).
Next, we can calculate the moles of nitrogen and ammonia produced in the reaction:
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Calculate the moles of N₂ used: \[ \text{Moles of } N_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 , \text{g}}{28.0 , \text{g/mol}} \approx 0.357 , \text{mol} \]
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According to the balanced equation \( N_2 + 3H_2 \rightarrow 2NH_3 \), 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
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Calculate the moles of NH₃ produced: \[ \text{Moles of } NH_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0 , \text{g}}{17.0 , \text{g/mol}} \approx 0.706 , \text{mol} \]
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From the stoichiometry of the reaction:
- From 1 mole of N₂, we produce 2 moles of NH₃.
- Thus, 0.357 moles of N₂ will produce: \[ \text{Moles of } NH_3 = 0.357 , \text{mol} \times 2 = 0.714 , \text{mol} \]
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Since \( 0.706 , \text{mol} \) of NH₃ is produced, this is consistent with our earlier calculation (indicating possibly some minor losses or differences in theoretical versus actual yields).
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Next, we can determine the moles of H₂ needed:
- From the balanced equation, 1 mole of N₂ requires 3 moles of H₂.
- Therefore, \( 0.357 \) moles of N₂ would require: \[ 0.357 , \text{mol} \times 3 = 1.071 , \text{mol} , H_2 \]
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Calculate the mass of H₂ used: \[ \text{Mass of } H_2 = \text{moles} \times \text{molar mass} = 1.071 , \text{mol} \times 2.0 , \text{g/mol} \approx 2.14 , \text{g} \]
Therefore, rounding the value, the mass of hydrogen used in the reaction is approximately:
2.0 g of H₂