Question #934b0

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Question #934b0

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Answer 1 · 2016-12-16T02:46:08

We need (i) a stoichiometrically balanced equation:

$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} (g) + 5 H_{2} O (g)$ $C_4H_10 + 13/2O_2 rarr 4CO_2(g) + 5H_2O(g)$

Explanation:

And then (ii) equiv quantities of reactant and product:

$Moles of butane$ $"Moles of butane"$ $=$ $=$ $\frac{5.80 \cdot g}{58.12 \cdot g \cdot m o l^{- 1}} = 0.0998 \cdot m o l$ $(5.80*g)/(58.12*g*mol^-1)=0.0998*mol$

$Moles of dioxygen = \frac{1 \cdot a t m \times 15.0 \cdot L}{0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 273.15 K} = 0.669 \cdot m o l$ $"Moles of dioxygen"=(1*atmxx15.0*L)/(0.0821*L*atm*K^-1*mol^-1xx273.15K)=0.669*mol$

(This is not quite $STP$ $"STP"$ , but you will have to adapt this expression.) At any rate there is stoichiometric dioxygen for complete combustion, which we must presume.

Given the equation (i), $4 \times 0.0998 \cdot m o l$ $4xx0.0998*mol$ $C O_{2}$ $CO_2$ are evolved per equiv butane.

$V = \frac{n R T}{P}$ $V=(nRT)/P$

$= \frac{4 \times 0.0998 \cdot m o l \times \frac{0.0821 \cdot L \cdot a t m}{K \cdot m o l} \times 273.15 K}{1 \cdot a t m} = 8.95 L$ $=(4xx0.0998*molxx(0.0821*L*atm)/(K*mol)xx273.15K)/(1*atm)=8.95L$

And thus approx. $9 \cdot L$ $9*L$ carbon dioxide gas evolve from complete combustion.

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Question #934b0

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