Question

A reaction between liquid reactants takes place at −10.0°C in a sealed, evacuated vessel with a measured volume of 45.0L. Measurements show that the reaction produced 19.g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride 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.

Answer 1

Answer: 0.0624 atm

Explanation:-

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = 45.0 L

T= Temperature of the gas = -10.0°C = 263 K    $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

Moles of gas=$\frac{\text{ given mass}}{\text{ molar mass}}= \frac{19.0g}{146g/mole}=0.130moles$

$P=\frac{nRT}{V}=\frac{0.130\times 0.0821\times 263}{45.0}=0.0624atm$

The pressure of sulfur hexafluoride gas in the reaction vessel after the reaction is 0.0624 atm