Question

a reaction between liquid reactants takes place at 5.0 °c in a sealed, evacuated vessel with a measured volume of 35.0 l. measurements show that the reaction produced 31. g of dinitrogen monoxide gas. calculate the pressure of dinitrogen monoxide gas in the reaction vessel after the reaction. you may ignore the volume of the liquid reactants. be sure your answer has the correct number of significant digits. pressure: atm

Explanation:

Step1: Calculate moles of NO

Molar mass of NO ($M$) is $14.01 + 16.00 = 30.01\ \text{g/mol}$. Moles ($n$) = $\frac{\text{mass}}{\text{molar mass}}=\frac{31\ \text{g}}{30.01\ \text{g/mol}}\approx1.033\ \text{mol}$.

Step2: Convert temperature to Kelvin

$T = 5.0^\circ\text{C}+ 273.15 = 278.15\ \text{K}$. Volume $V = 35.0\ \text{L}$.

Step3: Apply Ideal Gas Law ($PV = nRT$)

$R = 0.0821\ \text{L·atm/(mol·K)}$. Solve for $P$: $P=\frac{nRT}{V}$.
Substitute values: $P=\frac{1.033\ \text{mol}\times0.0821\ \text{L·atm/(mol·K)}\times278.15\ \text{K}}{35.0\ \text{L}}$.
Calculate numerator: $1.033\times0.0821\times278.15\approx1.033\times22.83\approx23.69$.
Then $P=\frac{23.69}{35.0}\approx0.677\ \text{atm}$.

Answer:

$\boxed{0.68}$ (rounded to two significant figures, or 0.677 as precise; adjust based on sig figs: 31 g has two, 35.0 L three, 5.0 °C two. So two sig figs: 0.68 atm)