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39 grams of an alloy of aluminium and magnesium when heated with excess of dilute hydrochloric acid form magnesium chloride , aluminium chloride and hydrogen. The evolved hydrogen collected at 0 degree centigrade has a volume of 44.8 litre at 1 atmospheric pressure calculate the composition of alloy by moles and there respected percentage
let the mole of magnesium be x and mole of aluminium be yMg + 2Hcl---MgCL2 +H2Al +3Hcl ----AlCl3 +1.5H2NOW24x +27y=39x +1.5y=2solving these two equation we getx=.5 ,y=1percentage of magnesium=30.76%percentage of lauminium=69.24%
applying PV=nRT for hydrogenn=2 molelet the mole of magnesium be x and mole of aluminium be yMg + 2Hcl---MgCL2 +H2Al +3Hcl ----AlCl3 +1.5H2NOW24x +27y=39x +1.5y=2solving these two equation we get x=.5 ,y=1 percentage of magnesium=30.76% percentage of lauminium=69.24%
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