Sure, let’s break down the problem step-by-step to find the theoretical yield of [tex]\( \text{NO}_2 \)[/tex].
1. Determine the molar mass of the reactant [tex]\( \text{NO} \)[/tex]:
- The molar mass of [tex]\( \text{NO} \)[/tex] is 30.01 g/mol.
2. Convert grams of [tex]\( \text{NO} \)[/tex] to moles:
- Given mass of [tex]\( \text{NO} \)[/tex] is 25.0 grams.
- Use the molar mass to find the number of moles of [tex]\( \text{NO} \)[/tex]:
[tex]\[
\text{Moles of NO} = \frac{\text{Mass of NO}}{\text{Molar mass of NO}} = \frac{25.0 \text{ grams}}{30.01 \text{ g/mol}} \approx 0.833 \text{ moles}
\][/tex]
3. Use the balanced chemical equation to determine moles of [tex]\( \text{NO}_2 \)[/tex]:
- The chemical equation is:
[tex]\[
2 \text{NO} (g) + \text{O}_2 (g) \rightarrow 2 \text{NO}_2 (g)
\][/tex]
- From the equation, 2 moles of [tex]\( \text{NO} \)[/tex] react to form 2 moles of [tex]\( \text{NO}_2 \)[/tex].
- Therefore, the mole ratio of [tex]\( \text{NO} \)[/tex] to [tex]\( \text{NO}_2 \)[/tex] is 1:1.
- This means the moles of [tex]\( \text{NO} \)[/tex] will be equal to the moles of [tex]\( \text{NO}_2 \)[/tex]:
[tex]\[
\text{Moles of NO}_2 \approx 0.833 \text{ moles}
\][/tex]
4. Determine the molar mass of [tex]\( \text{NO}_2 \)[/tex]:
- The molar mass of [tex]\( \text{NO}_2 \)[/tex] is 46.01 g/mol.
5. Calculate the theoretical yield in grams of [tex]\( \text{NO}_2 \)[/tex]:
- Use the moles of [tex]\( \text{NO}_2 \)[/tex] and its molar mass to find the mass:
[tex]\[
\text{Theoretical yield of NO}_2 = \text{Moles of NO}_2 \times \text{Molar mass of NO}_2 \approx 0.833 \text{ moles} \times 46.01 \text{ g/mol} \approx 38.33 \text{ grams}
\][/tex]
So, the theoretical yield of [tex]\(NO _2\)[/tex] is approximately [tex]\(38.33\)[/tex] grams.
