To determine how much urea \((NH_2)_2CO\) can be produced from the reaction between ammonia \(NH_3\) and carbon dioxide \(CO_2\), we first need to write the balanced chemical equation for the reaction:
\[ 2 NH_3 + CO_2 \rightarrow (NH_2)_2CO + H_2O \]
From the balanced equation, we see that 2 moles of ammonia react with 1 mole of carbon dioxide to produce 1 mole of urea and 1 mole of water.
Next, we need to calculate the number of moles of \(NH_3\) and \(CO_2\) given the masses provided.
Step 1: Calculate moles of \(NH_3\) and \(CO_2\)
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Molar mass of \(NH_3\):
- \(N: 14.01 , g/mol\)
- \(H: 1.008 , g/mol\) (3 hydrogens)
\[ \text{Molar mass of } NH_3 = 14.01 + (3 \times 1.008) = 17.034 , g/mol \]
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Moles of \(NH_3\): \[ \text{Moles of } NH_3 = \frac{51 , g}{17.034 , g/mol} \approx 3.00 , mol \]
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Molar mass of \(CO_2\):
- \(C: 12.01 , g/mol\)
- \(O: 16.00 , g/mol\) (2 oxygens)
\[ \text{Molar mass of } CO_2 = 12.01 + (2 \times 16.00) = 44.01 , g/mol \]
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Moles of \(CO_2\): \[ \text{Moles of } CO_2 = \frac{176 , g}{44.01 , g/mol} \approx 4.00 , mol \]
Step 2: Determine the limiting reactant
According to the balanced equation, 2 moles of \(NH_3\) react with 1 mole of \(CO_2\).
- From 3 moles of \(NH_3\):
- Required moles of \(CO_2\) = \(\frac{3 , mol , NH_3}{2} = 1.5 , mol , CO_2\)
We have 4 moles of \(CO_2\) available, which is more than enough. Hence, \(NH_3\) is the limiting reactant.
Step 3: Calculate the maximum amount of urea produced
From the reaction, 2 moles of \(NH_3\) produce 1 mole of urea.
- Therefore, 3 moles of \(NH_3\) will produce: \[ \text{Moles of urea} = \frac{3 , mol , NH_3}{2} = 1.50 , mol , urea \]
Step 4: Calculate the mass of urea produced
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Molar mass of urea \((NH_2)_2CO\):
- \(N: 14.01 , g/mol\) (2 nitrogens)
- \(H: 1.008 , g/mol\) (4 hydrogens)
- \(C: 12.01 , g/mol\)
- \(O: 16.00 , g/mol\)
\[ \text{Molar mass of urea} = (2 \times 14.01) + (4 \times 1.008) + 12.01 + 16.00 = 60.06 , g/mol \]
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Mass of urea produced: \[ \text{Mass of urea} = 1.50 , mol \times 60.06 , g/mol \approx 90.09 , g \]
In conclusion, the maximum amount of urea that can be prepared from the reaction of 51g \(NH_3\) with 176g \(CO_2\) is approximately 90.09 grams.