Question

If 0.291 g of Mg is reacted with excess HCl and 301 mL of H2 gas is collected in a gas collection tube over water at 25oC and 1 atm of pressure. What is the experimental partial pressure of water vapor in the hydrogen gas?

Answer 1

Solution :-

Balanced reaction equation

Mg + 2HCl ----- > MgCl2 + H2

now lets calculate the total presusre in the gas collection tube

lets first calculate the moles of the Mg metal

0.291 g Mg * 1 mol 24.305 g = 0.012 mol Mg

1 mol Mg = 1 mol H2

so moles of H2 produced = 0.012 mol

temeprature = 25 C +273 = 298 K

Volume = 301 ml = 0.301 L

PV= nRT

P = nRT/V

   = 0.012 mol * 0.08206 L atm per mol K * 298 K / 0.301 L

= 0.9749 atm

so the total pressure = 0.9749 atm

total pressure = 1 atm

so the partial pressure of vapor of water = total pressure - partial pressure of H2

                                                                   = 1 atm -0.9749 atm

                                                                  = 0.0251 atm

So the vapor presusre of water = 0.0251 atm.