A 0.250 g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H_2(g) is collected over water at 29 degrees C and 752 torr, the volume is found to be 305 mL. The vapor pressure of water at 29 degrees C is 30.0 torr.What is the mass percentage of aluminum in this alloy?
The question I got is really similar to this one except that the volume is 295 mL rather than 305 mL. The correct answer for my problem is 28.3%. I followed the steps A.S. put up, and it's correct!! I attached a picture of my problem from Mastering Chemistry, which has the correct answer with a green box around it.
let there be x gram of magnesium and y gram of aluminium in the magnesium-aluminium alloy so x + y = 0.25 ....(1) now reaction of magnesium with HCl... Mg + 2HCl ------ MgCl2 + H2.......(2) reaction of aluminium with HCl ... 2Al + 6HCl ------- 2AlCl3 + 3H2.........(3) so H2 is derived from both Al and Mg reactions with HCl.... calculating no. of moles of H2 produced from both reactions... atomic mass of Mg = 24.3 g/mole and atomic mass of Al = 27 g/mole from reaction (2)....we can say that 24.3 g of Mg produces 1 mole of H2 so x g of Mg will produce 1/24.3 X x = x/24.3 moles of H2 similarly from reaction (3)...we can say that 2 X 27 = 54 g of Al produces 3 moles of H2 so y g of Al will produce 3/54 X y = 3y/54 moles of H2 so total number of moes of H2 produced = 3y/54 + x/24.3 now calculating number of moles of H2 produced with the help of following equation... PV = nRT P (due to hydrogen) = total pressure - pressure due to water vapour = 752 - 30 = 722 torr = 722/760 = 0.95 atm (1 atm = 760 torr) V = 305 ml or 305/1000 = 0.305 L R = 0.0821 L atm /K/mole T = 29 + 273 = 302 K n = ? putting the values... 0.95 X 0.305 = n X 0.0821 X 302 n = 0.289/24.794 = 0.012 as total no. of moles = 3y/54 + x/24.3 .... so 3y/54 + x/24.3 = 0.012 .....(4) now we have two equations in x and y ....i.e. (1) and (4) which can be solved from eqn(4)... x/24.3 = 0.012 - 3y/54 x/24.3 = (0.012 X 54 - 3y)/54 x/24.3 = (0.648 - 3y) / 54 x = 24.3 X (0.648-3y)/54 = 0.45 (0.648 - 3y) putting this value of x in eqn (1) 0.45(0.648-3y) + y = 0.25 0.292 - 1.35y + y = 0.25 0.292 -0.35y = 0.25 0.292 - 0.25 = 0.35 y y = 0.042/0.35 = 0.12 g so x = 0.25-0.12 = 0.13 g so weight of Al = 0.12 g weight of Mg = 0.13 g mass % of Al = 0.12/0.25 X 100 = 48% please check the maths... also feel free to ask any questions..
