Answer to Question #304875 in General Chemistry for Ace

Question

$$\ce{2C_4H_{10}(g) +13O_2(g) -> 8 CO_2(g) + 10H_2O(l)}$$

$65cm^3$ of butane is burnt in $65cm^3$ of oxygen. Find the volume of carbon dioxide produced.\

Solution\

Assume\

i) $O_2$ and $CO_2$ are ideal gas at STP condition $22.7dm^3/mol$\

ii) $C_4H_{10}$ is limiting reactant with lowest coefficient in the equation while $O_2$ is the excess reactant.

$n_{O_2 (real)}=0.065/22.7=2.8634\times10^-3$ mol $O_2$(real)

$$From\qquad equation:$$$$\ce{2C_4H_{10}(g) :13O_2(g) }$$

$$Real:$$$$\ce{X C_4H_{10}(g) :(2.8634\times{10^-3}) O_2(g) }$$

Ratio: $\frac{2}{13}=\frac{X}{2.8634\times{10^-3}}$\

$X=\frac{2}{13}\times(2.8634\times{10^-3})=4.4052\times10^-4$ mol $C_4H_{10}(real)$ which is the limiting reactant.

$$From\qquad equation:$$$$\ce{2C_4H_{10}(g) :8CO_2(g) }$$

$$Real:$$$$\ce{4.4052\times10^{-4} C_4H_{10}(g) :Y CO_2(g) }$$

$Y=\frac{8}{2}\times(4.4052\times10^{-4})=1.7621\times10^-3$ mol $CO_2(real)$ 

$V_{CO_2}(real)=1.7621\times10^-3$ mol $\times 22.7dm^3/mol=0.039999dm^3=0.04dm^3=40cm^3$

Correct?

Answer 1

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