To find the maximum mass of NO that can be obtained from 15.5 grams of [tex]\( N_2O_4 \)[/tex] and 4.68 grams of [tex]\( N_2H_4 \)[/tex], follow these steps:
1. Determine the molar masses of the reactants and product:
[tex]\[
\text{Molar mass of } N_2O_4 = 92.02 \, \text{g/mol}
\][/tex]
[tex]\[
\text{Molar mass of } N_2H_4 = 32.05 \, \text{g/mol}
\][/tex]
[tex]\[
\text{Molar mass of } NO = 30.01 \, \text{g/mol}
\][/tex]
2. Calculate the number of moles of each reactant:
[tex]\[
\text{Moles of } N_2O_4 = \frac{15.5 \, \text{g}}{92.02 \, \text{g/mol}} = 0.1684 \, \text{moles}
\][/tex]
[tex]\[
\text{Moles of } N_2H_4 = \frac{4.68 \, \text{g}}{32.05 \, \text{g/mol}} = 0.1460 \, \text{moles}
\][/tex]
3. Using the balanced chemical equation, identify the stoichiometry:
[tex]\(2 \, \text{moles of } N_2O_4\)[/tex] react with [tex]\(1 \, \text{mole of } N_2H_4\)[/tex] to produce [tex]\(6 \, \text{moles of } NO\)[/tex].
4. Determine the limiting reactant:
- Calculate the moles of NO that each reactant can produce:
[tex]\[
\text{From } N_2O_4: \frac{6}{2} \times 0.1684 \, \text{moles} = 0.5051 \, \text{moles of } NO
\][/tex]
[tex]\[
\text{From } N_2H_4: \frac{6}{1} \times 0.1460 \, \text{moles} = 0.8760 \, \text{moles of } NO
\][/tex]
- The limiting reactant is the one that produces the fewer moles of NO. Therefore, [tex]\( N_2O_4 \)[/tex] is the limiting reactant since 0.5051 moles is less than 0.8760 moles.
5. Calculate the mass of NO produced by the limiting reactant:
[tex]\[
\text{Moles of NO produced} = 0.5051 \, \text{moles}
\][/tex]
[tex]\[
\text{Mass of NO produced} = 0.5051 \, \text{moles} \times 30.01 \, \text{g/mol} = 15.1648 \, \text{grams}
\][/tex]
So the maximum mass of NO that can be obtained from 15.5 grams of [tex]\( N_2O_4 \)[/tex] and 4.68 grams of [tex]\( N_2H_4 \)[/tex] is 15.1648 grams.
