Question

use the balanced equation, $2\ce{no} + \ce{o_{2}} \
ightarrow 2\ce{no_{2}}$, to answer the following question. how many grams of $\ce{o_{2}}$ are needed to react completely with $12.8\\ \text{g}$ of $\ce{no}$? be sure your answer has the correct number of significant figures.

Explanation:

Step1: Calculate moles of NO

Molar mass of NO: $M(\text{NO}) = 14.01 + 16.00 = 30.01\ \text{g/mol}$.
Moles of NO: $n(\text{NO}) = \frac{12.8\ \text{g}}{30.01\ \text{g/mol}} \approx 0.4265\ \text{mol}$.

Step2: Relate moles of NO to O₂

From balanced equation $2\text{NO} + \text{O}_2
ightarrow 2\text{NO}_2$, mole ratio $\text{NO}:\text{O}_2 = 2:1$.
Moles of O₂: $n(\text{O}_2) = \frac{0.4265\ \text{mol}}{2} \approx 0.21325\ \text{mol}$.

Step3: Calculate mass of O₂

Molar mass of O₂: $M(\text{O}_2) = 32.00\ \text{g/mol}$.
Mass of O₂: $m(\text{O}_2) = 0.21325\ \text{mol} \times 32.00\ \text{g/mol} \approx 6.824\ \text{g}$.

Answer:

6.82 (or 6.8 g, depending on significant figures; 12.8 has 3 sig figs, so 6.82 g)