To determine how many grams of ammonium nitrate (NH4NO3) are required to produce 12.0 grams of dinitrogen monoxide (N2O), we need to follow several steps involving stoichiometry.
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Write the balanced chemical equation: \[ 2 , \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2, \text{H}_2\text{O} \]
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Determine the molar mass of N2O:
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol
\[ \text{Molar mass of } \text{N}_2\text{O} = (2 \times 14.01) + (1 \times 16.00) = 28.02 + 16.00 = 44.02 , \text{g/mol} \]
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Convert grams of N2O to moles: \[ \text{Moles of N}_2\text{O} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{12.0 , \text{g}}{44.02 , \text{g/mol}} \approx 0.272 , \text{mol} \]
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Use the stoichiometry from the balanced equation: From the balanced equation, 1 mole of N2O is produced from 2 moles of NH4NO3. Therefore:
\[ \text{Moles of NH}_4\text{NO}_3 = 2 \times \text{Moles of N}_2\text{O} = 2 \times 0.272 , \text{mol} \approx 0.544 , \text{mol} \]
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Determine the molar mass of NH4NO3:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
\[ \text{Molar mass of } \text{NH}_4\text{NO}_3 = (2 \times 14.01) + (4 \times 1.01) + (3 \times 16.00) \] \[ = 28.02 + 4.04 + 48.00 = 80.06 , \text{g/mol} \]
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Convert moles of NH4NO3 to grams: \[ \text{Mass of NH}_4\text{NO}_3 = \text{Moles} \times \text{Molar mass} = 0.544 , \text{mol} \times 80.06 , \text{g/mol} \approx 43.55 , \text{g} \]
Thus, approximately 43.55 grams of ammonium nitrate are needed to produce 12.0 grams of N2O.
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