To determine how many grams of nitrogen were present initially when producing a certain amount of ammonia, we can follow these steps:
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Write down the balanced chemical equation: \[ N_2 + 3H_2 \rightarrow 2NH_3 \]
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Determine the molar masses:
- Molar mass of \(N_2\) (for nitrogen) = \(2 \times 14 , \text{g/mol} = 28 , \text{g/mol}\)
- Molar mass of \(H_2\) (for hydrogen) = \(2 \times 1 , \text{g/mol} = 2 , \text{g/mol}\)
- Molar mass of \(NH_3\) (for ammonia) = \(14 , \text{g/mol} + 3 \times 1 , \text{g/mol} = 17 , \text{g/mol}\)
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Examine the stoichiometry: According to the balanced equation:
- 1 mole of \(N_2\) produces 2 moles of \(NH_3\).
- 1 mole of \(N_2\) corresponds to \(28 , \text{g}\) of nitrogen and produces \(34 , \text{g}\) of \(NH_3\) (since \(2 \times 17 , \text{g/mol} = 34 , \text{g}\)).
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Calculate the mass of nitrogen based on the mass of ammonia produced: If \(x\) grams of ammonia is produced, we can set up a proportion based on the stoichiometric ratios: \[ \frac{28 , \text{g} , N_2}{34 , \text{g} , NH_3} = \frac{m , \text{g} , N_2}{x , \text{g} , NH_3} \] Rearranging this gives: \[ m = \frac{28 , \text{g}}{34 , \text{g}} \cdot x \]
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Calculate the amount of nitrogen for the specific amount of ammonia:
- If \(x = 68 , \text{g}\) (assuming the amount of ammonia produced is given as 68 g), we can substitute \(x\): \[ m = \frac{28}{34} \cdot 68 \] \[ m = \frac{28 \times 68}{34} = 56 , \text{grams of } N_2 \]
Thus, if 68 grams of ammonia were produced, there were initially 56 grams of nitrogen present.
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