5ml of gas containing only carbon and hydrogen were mixed with an excess of oxygen (30 ml) and the mixture exploded by means of electric spark. After the explosion the mixture of remaining gas will be 25 ml. On adding a concentrated solution of KOH the volume further reduce to 15 ml, the residual gas being pure oxygen. All volume have been reduce to NTP. Calculate the molecular formula of hydrogen gas.
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Volume of oxygen taken = 30 ml
Volume of unused oxygen = 15 ml
Volume of O2 used = Volume of O2 added – Volume of O2 left
= 30 – 15 = 15 ml
Volume of CO2 produced
= Volume of gaseous mixture after explosion – Volume of unused oxygen
Or Volume of CO2produced = 25 – 15 = 10 ml
Volume of hydrocarbon = 5 m;
General equation for combustion of a hydrocarbon is as follows –
CxHy + (x + y/4)
O2 → xCO2 +y/4 H2O
(Hydrocarbon)
5 ml 5(x + y/4)ml 5 x
∴ Volume of CO2 produced = 5x, Since Volume of CO2 = 10 ml
∴ 5x = ⇒ x = 2, Volume of O2 used = ml
∴ 5 (x + y/4) = 15 ⇒ x + y/4 = 3
⇒ 2 + y/4 = 3 (∵ x = 2)
⇒ 8 + y = 12 ∴ y = 4
Volume of unused oxygen = 15 ml
Volume of O2 used = Volume of O2 added – Volume of O2 left
= 30 – 15 = 15 ml
Volume of CO2 produced
= Volume of gaseous mixture after explosion – Volume of unused oxygen
Or Volume of CO2produced = 25 – 15 = 10 ml
Volume of hydrocarbon = 5 m;
General equation for combustion of a hydrocarbon is as follows –
CxHy + (x + y/4)
O2 → xCO2 +y/4 H2O
(Hydrocarbon)
5 ml 5(x + y/4)ml 5 x
∴ Volume of CO2 produced = 5x, Since Volume of CO2 = 10 ml
∴ 5x = ⇒ x = 2, Volume of O2 used = ml
∴ 5 (x + y/4) = 15 ⇒ x + y/4 = 3
⇒ 2 + y/4 = 3 (∵ x = 2)
⇒ 8 + y = 12 ∴ y = 4
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