39 gm of an alloy of aluminium and magnesium when heated with excess of dil. HCl, forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected at 0°C has a volume of 44.8L at 1ATM pressure. Calculate composition of alloy by moles.
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applying PV=nRT for hydrogen
n=2 mole
let the mole of magnesium be x and mole of aluminium be y
Mg + 2Hcl---MgCL2 +H2
Al +3Hcl ----AlCl3 +1.5H2
NOW
24x +27y=39
x +1.5y=2
solving these two equation we get x=.5 ,y=1 percentage of magnesium=30.76% percentage of lauminium=69.24%
sparshhh19:
I've seen this solution in many places but the answer is wrong. The answer should be 66.6% aluminium.
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