To determine how many grams of nitrogen were present initially when 68 grams of ammonia (NH₃) were produced, we can use stoichiometry based on the balanced chemical reaction:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
From the equation, we can see that 2 moles of ammonia (NH₃) are produced from 1 mole of nitrogen (N₂). We will follow these steps:
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Calculate the molar mass of ammonia (NH₃):
- Nitrogen (N) has an atomic mass of 14 g/mol.
- Hydrogen (H) has an atomic mass of 1 g/mol.
- Therefore, the molar mass of NH₃ is: \[ 14 , \text{g/mol (N)} + 3 \times 1 , \text{g/mol (H)} = 14 + 3 = 17 , \text{g/mol} \]
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Determine the moles of ammonia produced:
- Given that 68 grams of NH₃ were produced: \[ \text{Moles of NH₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{68 , \text{g}}{17 , \text{g/mol}} = 4 , \text{moles} \]
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Use stoichiometry to find moles of nitrogen used:
- The reaction shows that 2 moles of NH₃ are produced from 1 mole of N₂, so for every 2 moles of NH₃, we consume 1 mole of N₂: \[ \text{Moles of N₂} = \frac{4 , \text{moles NH₃}}{2} = 2 , \text{moles N₂} \]
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Calculate the mass of nitrogen used:
- Molar mass of nitrogen (N₂) is: \[ 2 \times 14 , \text{g/mol} = 28 , \text{g/mol} \]
- The mass of nitrogen used is: \[ \text{mass of N₂} = \text{moles} \times \text{molar mass} = 2 , \text{moles} \times 28 , \text{g/mol} = 56 , \text{grams} \]
Therefore, the amount of nitrogen present initially was 56 grams.
The correct answer is:
56