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

Jul 4, 2017

$V o l u m e o f C {O}_{2} = 11.25 {\mathrm{dm}}^{3}$

#### Explanation:

Here is the equation of reaction (Combustion of Carbon)

$C {O}_{2} \to C + {O}_{2}$

Mass of solute $= 6 g$

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

Recall $\to$ Amount in moles $= \frac{M a s s}{M o l a r m a s s}$

$\text{Amount in moles} = \frac{M a s s o f C a r b o n}{M o l a r m a s s o f c a r b o n}$

$\text{Amount in moles} = \frac{6}{12}$

$\text{Amount in moles of C} = 0.5 m o l s$

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

Since $1 m o l$ of $C {O}_{2}$ yields $1 m o l$ of $C$

$\therefore$ $0.5 m o l$ of $C {O}_{2}$ yields $0.5 m o l$ of $C$

Hence $\text{No of moles of} C {O}_{2} = 0.5 m o l s$

Also Recall $\to$ Amount in moles $= \text{Volume of solution"/"Molar volume}$

$\to$ $\text{Volume of" CO_2 = "Number of moles of" CO_2 xx "Molar Volume}$

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

$\text{Volume of} C {O}_{2} = 0.5 \cancel{m o l} \times 22.5 \cancel{m o {l}^{-} 1} {\mathrm{dm}}^{3}$

$\text{Volume of} C {O}_{2} = \left(0.5 \times 22.5\right) {\mathrm{dm}}^{3}$

$V o l u m e o f C {O}_{2} = 11.25 {\mathrm{dm}}^{3}$