To determine how many grams of dinitrogen pentoxide (N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex]) are required to produce 37.18 grams of nitric acid (HNO[tex]\(_3\)[/tex]), we will follow a step-by-step approach:
1. Determine the molar masses:
- The molar mass of N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex] is 108.01 g/mol.
- The molar mass of HNO[tex]\(_3\)[/tex] is 63.01 g/mol.
2. Calculate the number of moles of HNO[tex]\(_3\)[/tex] produced:
- We start with 37.18 grams of HNO[tex]\(_3\)[/tex].
- To find the number of moles, use the formula:
[tex]\[
\text{moles of HNO}_3 = \frac{\text{mass of HNO}_3}{\text{molar mass of HNO}_3}
\][/tex]
Substituting the given values:
[tex]\[
\text{moles of HNO}_3 = \frac{37.18 \, \text{g}}{63.01 \, \text{g/mol}} \approx 0.5901 \, \text{mol}
\][/tex]
3. Relate moles of HNO[tex]\(_3\)[/tex] to moles of N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex]:
- According to the balanced chemical equation, 1 mole of N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex] produces 2 moles of HNO[tex]\(_3\)[/tex].
- Thus, to find the moles of N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex] required, we divide the moles of HNO[tex]\(_3\)[/tex] by 2:
[tex]\[
\text{moles of N}_2\text{O}_5 = \frac{\text{moles of HNO}_3}{2}
\][/tex]
Substituting the calculated moles of HNO[tex]\(_3\)[/tex]:
[tex]\[
\text{moles of N}_2\text{O}_5 = \frac{0.5901 \, \text{mol}}{2} \approx 0.2950 \, \text{mol}
\][/tex]
4. Calculate the required mass of N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex]:
- To find the mass of N[tex]\(_2\)[/tex]O[tex]\(_5\)[/tex], use the formula:
[tex]\[
\text{mass of N}_2\text{O}_5 = \text{moles of N}_2\text{O}_5 \times \text{molar mass of N}_2\text{O}_5
\][/tex]
Substituting the values:
[tex]\[
\text{mass of N}_2\text{O}_5 = 0.2950 \, \text{mol} \times 108.01 \, \text{g/mol} \approx 31.87 \, \text{g}
\][/tex]
Therefore, 31.87 grams of dinitrogen pentoxide are required to produce 37.18 grams of nitric acid.
