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Question Number 18429 by tawa tawa last updated on 21/Jul/17

30cm^3 of hydrogen at s.t.p combines with 20cm^3 of oxygen to form steam according to the following equation, 2H_2 (g) + O_2 (g) → 2H_2 O (g). Calculate the total volume of gaseous mixture at the end of the reaction.

$$\mathrm{30cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{hydrogen}\:\mathrm{at}\:\mathrm{s}.\mathrm{t}.\mathrm{p}\:\mathrm{combines}\:\mathrm{with}\:\mathrm{20cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{oxygen}\:\mathrm{to}\:\mathrm{form}\:\mathrm{steam}\: \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation},\:\:\mathrm{2H}_{\mathrm{2}} \:\left(\mathrm{g}\right)\:+\:\mathrm{O}_{\mathrm{2}} \:\left(\mathrm{g}\right)\:\rightarrow\:\mathrm{2H}_{\mathrm{2}} \mathrm{O}\:\left(\mathrm{g}\right). \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{total}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{gaseous}\:\mathrm{mixture}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{reaction}. \\ $$

Commented by tawa tawa last updated on 21/Jul/17

Please help with workings

$$\mathrm{Please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{workings} \\ $$

Commented by tawa tawa last updated on 21/Jul/17

I don′t know it sir

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{it}\:\mathrm{sir} \\ $$

Commented by tawa tawa last updated on 21/Jul/17

Help me with the way you think it should be. God bless you sir.

$$\mathrm{Help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{the}\:\mathrm{way}\:\mathrm{you}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tinkutara last updated on 21/Jul/17

1 cm^3 of O_2 requires 2 cm^3 of H_2 . ∴ 20 cm^3 of O_2 requires 40 cm^3 of H_2 . But available H_2 is only 30 cm^3 . Hence it is the limiting reagent. Now 2 cm^3 of H_2 produces 2 cm^3 of H_2 O. ∴ 30 cm^3 of H_2 produces 30 cm^3 of H_2 O. Volume of O_2 left unreacted = 5 cm^3 Hence final volume after the reaction of is 35 cm^3 .

$$\mathrm{1}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{O}_{\mathrm{2}} \:\mathrm{requires}\:\mathrm{2}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} . \\ $$$$\therefore\:\mathrm{20}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{O}_{\mathrm{2}} \:\mathrm{requires}\:\mathrm{40}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} . \\ $$$$\mathrm{But}\:\mathrm{available}\:\mathrm{H}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{only}\:\mathrm{30}\:\mathrm{cm}^{\mathrm{3}} .\:\mathrm{Hence} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{limiting}\:\mathrm{reagent}. \\ $$$$\mathrm{Now}\:\mathrm{2}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \:\mathrm{produces}\:\mathrm{2}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \mathrm{O}. \\ $$$$\therefore\:\mathrm{30}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \:\mathrm{produces}\:\mathrm{30}\:\mathrm{cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \mathrm{O}. \\ $$$$\mathrm{Volume}\:\mathrm{of}\:\mathrm{O}_{\mathrm{2}} \:\mathrm{left}\:\mathrm{unreacted}\:=\:\mathrm{5}\:\mathrm{cm}^{\mathrm{3}} \\ $$$$\mathrm{Hence}\:\mathrm{final}\:\mathrm{volume}\:\mathrm{after}\:\mathrm{the}\:\mathrm{reaction} \\ $$$$\mathrm{of}\:\mathrm{is}\:\mathrm{35}\:\mathrm{cm}^{\mathrm{3}} . \\ $$

Commented by tawa tawa last updated on 21/Jul/17

God bless you sir.

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

Commented by tawa tawa last updated on 23/Jul/17

The answer is correct sir.

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{sir}. \\ $$

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