$2.0 g$ of $H X_{2}$ and $8.0 g$ of $O X_{2}$ are placed in a $10.0 L$ vessel at $300 K$ (no reaction). Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the total pressure in atm (two decimals) is

Numerical answer type, answer in atm, tolerance 0.02.

Answer: 3.08 atm

n_{H X_{2}} = 2.0/2 = 1.0

n_{O X_{2}} = 8.0/32 = 0.25

, total

1.25 mol

P = \frac{1.25 \times 0.0821 \times 300}{10.0} = 3.0788 \approx 3.08 atm