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# A volume of 100 ml of water gas containing some CO2 was mixed with 100 ml of oxygen and mixture exploded. The volume after an explosion was 100 ml. On introducing NaOH, the volume was reduced to 52.5 ml. If the volume ratio of CO, H2 and CO2 in the original sample is ab:cd:2, calculate the value of ′abcd′.

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Solution

## Let, Vol. of CO=x mlVol. of H2=y ml & Vol. of CO2=z mlGiven that,x+y+z=100 mlNow,CO+12O2⟶CO2x 12x x mlAgain,H2+12O2⟶H2Oy 12y Negligible volumeTotal O2 used =x2+y2=x+y2Total CO2=x+zNow, before and after reaction with oxygen,CO+H2+CO2+(x2+y2O2)=CO2+H2O+ unreacted O2x y z 100ml (x+z) Neg. 52.5ml=100−x+y2After explosion total volume =100ml(x+z)+100−(x+y2)=100 x+z=x+y22x+2z=x+yx=y−2z -------- Eqn. 1After the explosion, volume of CO2= Total volume − Oxygen left unreacted =100−52.5=47.5ml x+z=47.5mlx=(47.5−z) -------- Eqn. 2Substitute 2 in 147.5−z=y−2z y−z=47.5ml ------ Eqn. 3 Again, since x+y+z=100 (47.5−z)+y+z=100 ⇒y=52.5ml ⇒z=52.5−47.5=5ml⇒x=100−(52.5+5)=42.5 mlVCO:VH2:VCO242.52.5=52.52.5=52.517:21:21721

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