How do I calculate the partial pressure of HCl?

How do I find pressure of H2 alone in mmHg? ?

• The problem says: calculate the value of R and the molar volume of H2 at STP given the following results: 0.234 g of Mg reacts with excess HCl(aq) to produce 240.7 mL of H2 gas collected over the water at a total pressure of 758.3mmHg and the temperature of 22.8 C.

• Answer:

  Write the reaction: Mg + 2HCl ------------> MgCl2 + H2 Step 1 : Convert mass of Mg to moles (0.234g/24.305g/mol) = 0.0096 mol Step 2 : Take mol ratio of Mg to H2 From the equation, 1 mol Mg = 1 mol H2 0.0096 mol Mg = 0.0096 mol H2 Step 3 : Use Ideal Gas Equation PV = nRT R = PV/nT P = Pressure = 758.3 mmHg = 0.997 atm V = Volume = 240.7mL = 0.2407L n = 0.0096 moles T = Temperature = 22.8C = 295.95 K R = (0.997atm * 0.2407L) / (0.0096 moles * 295.95K) = 0.0844 L*atm*K^-1*mol^-1 (Actual R value is 0.0821. The difference might be due to change in significant figures during calculation) PART 2 : At STP, 1 mol of any gas (molar volume of any gas) would be 22.4 L To prove this use again PV = nRT V = nRT/P Use standard conditions, P =1.03 atm, T = 273.15K V = (1 * 0.0821 * 273.15) /1.03 = 22.4L