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Question Number 12801 by tawa last updated on 01/May/17

Answered by sandy_suhendra last updated on 01/May/17

1) molar gas = ((2.8)/(22.4)) = 0.125 mol molar mass of gas = (8/(0.125)) = 64 gram/mol vapour density = ((molar mass of gas)/(molar mass of H_2 )) = ((64)/2) =32 2) Mg + 2HCl → MgCl_2 + H_2 Mg = 0.48 gr = ((0.48)/(24)) = 0.02 mol HCl = (2/1) × 0.02 = 0.04 mol volume of HCl = ((0.04)/(0.25)) = 0.16 dm^3

$$\left.\mathrm{1}\right)\:\mathrm{molar}\:\mathrm{gas}\:=\:\frac{\mathrm{2}.\mathrm{8}}{\mathrm{22}.\mathrm{4}}\:=\:\mathrm{0}.\mathrm{125}\:\mathrm{mol} \\ $$$$\mathrm{molar}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{gas}\:=\:\frac{\mathrm{8}}{\mathrm{0}.\mathrm{125}}\:=\:\mathrm{64}\:\mathrm{gram}/\mathrm{mol} \\ $$$$\mathrm{vapour}\:\mathrm{density}\:=\:\frac{\mathrm{molar}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{gas}}{\mathrm{molar}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} }\:=\:\frac{\mathrm{64}}{\mathrm{2}}\:=\mathrm{32}\:\:\:\:\: \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Mg}\:+\:\mathrm{2HCl}\:\rightarrow\:\mathrm{MgCl}_{\mathrm{2}} \:+\:\mathrm{H}_{\mathrm{2}} \\ $$$$\mathrm{Mg}\:=\:\mathrm{0}.\mathrm{48}\:\mathrm{gr}\:=\:\frac{\mathrm{0}.\mathrm{48}}{\mathrm{24}}\:=\:\mathrm{0}.\mathrm{02}\:\mathrm{mol} \\ $$$$\mathrm{HCl}\:=\:\frac{\mathrm{2}}{\mathrm{1}}\:×\:\mathrm{0}.\mathrm{02}\:=\:\mathrm{0}.\mathrm{04}\:\mathrm{mol} \\ $$$$\mathrm{volume}\:\mathrm{of}\:\mathrm{HCl}\:=\:\frac{\mathrm{0}.\mathrm{04}}{\mathrm{0}.\mathrm{25}}\:=\:\mathrm{0}.\mathrm{16}\:\mathrm{dm}^{\mathrm{3}} \\ $$$$ \\ $$

Commented by tawa last updated on 01/May/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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