Question

530 mL Need 18°C and 1.04 ATM is mixed with 364 mL of SF6 at 18°C and 0.85 ATM in a 250 mL flask. Calculate the partial pressure of each gas

Answer 1

#77307 Chemistry, Other

530 mL Neon 18°C and 1.04 ATM is mixed with 364 mL of SF₆ at 18°C and 0.85 ATM in a 250 mL flask. Calculate the partial pressure of each gas.

Answer:

pV=nRT

R = 0.082 L atm K⁻¹ mol⁻¹

n (Ne) = pV/RT

$$n(Ne) = \frac{(1.04 \times 0.53)}{(0.082 \times (18 + 273))} = 0.02 \, \text{mol}$$ $$n(SF_6) = \frac{(0.85 \times 0.364)}{(0.082 \times (18 + 273))} = 0.01 \, \text{mol}$$

p = nRT/V

$$p(Ne) = \frac{0.02 \times (0.082 \times (18 + 273))}{0.25} = 1.9 \, \text{mol}$$ $$p(SF_6) = \frac{0.01 \times (0.082 \times (18 + 273))}{0.25} = 0.95 \, \text{mol}$$