Question

a mixture of 4.0 g h₂(g) and 10.0 g he(g) in a 4.3 l flask is maintained at 0 °c. what is the partial pressure of h₂ gas? 5.12 atm 6.00 atm 2.50 atm 10.4 atm

Explanation:

Step1: Calculate moles of \(H_2\)

Molar mass of \(H_2\) is \(2\space g/mol\). Moles (\(n\)) = mass / molar mass. So \(n_{H_2}=\frac{4.0\space g}{2\space g/mol}=2\space mol\).

Step2: Convert temperature to Kelvin

\(T = 0^\circ C + 273.15 = 273.15\space K\), volume \(V = 4.3\space L\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\).

Step3: Apply ideal gas law for \(H_2\)

Ideal gas law: \(PV = nRT\). Solve for \(P\): \(P=\frac{nRT}{V}\). Plug in values: \(P=\frac{2\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times273.15\space K}{4.3\space L}\approx5.12\space atm\).

Answer:

5.12 atm