The vol of CO2 released when 6g of carbon combusrs completed ?

Question

2413 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-04T08:23:00

$Volume of CO_2 = 11.25 dm^3$

Here is the equation of reaction (Combustion of Carbon)

$CO_2 -> C + O_2$

Mass of solute $= 6g$

Molar mass of Carbon $= 12gmol^-1$

Recall $->$ Amount in moles $= (Mass)/(Molar mass)$

$"Amount in moles" = (Mass of Carbon)/(Molar mass of carbon)$

$"Amount in moles" = 6/12$

$"Amount in moles of C" =0.5 mols$

From the stoichiometric equation we have $1 : 1 : 1$ as the ratio

Since $1 mol$ of $CO_2$ yields $1 mol$ of $C$

$:.$ $0.5 mol$ of $CO_2$ yields $0.5mol$ of $C$

Hence $"No of moles of" CO_2 = 0.5 mols$

Also Recall $->$ Amount in moles $= "Volume of solution"/"Molar volume"$

$->$ $"Volume of" CO_2 = "Number of moles of" CO_2 xx "Molar Volume"$

Note that: $"Molar Volume" = 22.5mol^-1dm^3 "at s.t.p"$

$"Volume of" CO_2 = 0.5cancel(mol) xx 22.5cancel(mol^-1)dm^3$

$"Volume of" CO_2 = (0.5 xx 22.5) dm^3$

$Volume of CO_2 = 11.25 dm^3$