What is the masspercentage of aluminum in this alloy? Solution If pressure = 752

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Explanation / Answer

If pressure = 752 and vP = 30.00 torr, partial pressure of H2 = 722torr. n = PV/RT = 722 x 0.309/62.36367 x 302 = 0.01185 moles of H2 Let mass of Mg in alloy = a grams, mass of Al in alloy = (0.250 -a) grams It atomic mass of Mg = 24.30 and Al = 26.98, then moles Mg =a/24.30 and moles Al = (0.250 - a)/26.98 Mg + 2HCl ----> MgCl2 + H2 1 mol    1mol a/24.30 mol    a/24.30 mol Al +   3HCl  ------> AlCl3    + 1.5H2 1 mole          1.5mole (0.250 - a)/26.98 mole       1.5 x (0.250 - a)/26.98 mole So (a/24.30 mol)+ 1.5 x (0.250 - a)/26.98 mole = 0.01185moles of H2 So (a/24.30 mol) + 0.375 - 1.5a)/26.98 mole = 0.01185 molesof H2 So just need to solve for a 26.98a + 24.3 (0.375 - 1.5a) = 0.01185 x 24.3 x 26.98 26,98a + 9.1125 - 36.45a = 7.769 -9.47a = - 1.3435 a = 0.142g = mass of Mg Therefore, mass of aluminum = 0.250 total -0.142 Al= 0.108 g therefore, 0.108 g Al/0.250 g alloy * 100%= 43.2%Aluminum

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