Question

A piece of solid magnesium is reacted with dilute hydrochloric acid to form hydrogen gas: Mg(s) + 2 HCl(aq) Æ MgCl2(aq) + H2(g) What volume of H2 is collected over water at 28 °C by reaction of 1.25 g of Mg with 50.0 mL of 0.10 M HCl? The barometer records an atmospheric pressure of 748 torr and the vapor pressure of water at this temperature is 28.35 torr.

I am not quite sure what to do with the vapor pressure of water here.

Thank you.

Answer 1

T = 28|C = 301 K

m = 1.25 g of Mg

mol of Mg = mass/MW = 1.25/24.305 = 0.051429 mol of Mg

V = 50 ml = 0.05 L

M = 0.1 HCl

mol of HCl = MV = (0.05*0.1) = 0.005 mol of HCl

P = 748 torr

P°vap = 28.35 torr

ratio is 1:2

0.005 mol of HCl react with 0.005/2 =0.0025 mol of MG to form 0.005 mol of H2

then

PV = nRT

V = nRT/P

n = 0.005 mol of H2

R = 0.082

T = 301 K

P of H2 = PT - P°vap = 748-28.35 = 719.65 torr

change P of torr to atm = 719.65/760 = 0.94690 atm

V = nRT/P

V = 0.005*0.082*301/(0.94690) = 0.130330 L

V = 130.3 ml