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

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Answer 1 · 2017-07-04T08:23:00

$V o l u m e o f C O_{2} = 11.25 d m^{3}$ $Volume of CO_2 = 11.25 dm^3$

Here is the equation of reaction (Combustion of Carbon)

$C O_{2} \to C + O_{2}$ $CO_2 -> C + O_2$

Mass of solute $= 6 g$ $= 6g$

Molar mass of Carbon $= 12 g m o l^{- 1}$ $= 12gmol^-1$

Recall $\to$ $->$ Amount in moles $= \frac{M a s s}{M o l a r m a s s}$ $= (Mass)/(Molar mass)$

$Amount in moles = \frac{M a s s o f C a r b o n}{M o l a r m a s s o f c a r b o n}$ $"Amount in moles" = (Mass of Carbon)/(Molar mass of carbon)$

$Amount in moles = \frac{6}{12}$ $"Amount in moles" = 6/12$

$Amount in moles of C = 0.5 m o l s$ $"Amount in moles of C" =0.5 mols$

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

Since $1 m o l$ $1 mol$ of $C O_{2}$ $CO_2$ yields $1 m o l$ $1 mol$ of $C$ $C$

$:.$ $0.5 mol$ 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: $"Molar Volume" = 22.5mol^-1dm^3 "at s.t.p"$

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

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

$Volume of CO_2 = 11.25 dm^3$