4.3 g of an alkane is burnt in sufficient ox | vedclass

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(C) The combustion reaction of an alkane is: $C_nH_{2n+2} + \frac{3n+1}{2} O_2 	o nCO_2 + (n+1) H_2O$The reaction of $CO_2$ with $NaOH$ to form $Na_2

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$C_6H_{14}$

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(C) The combustion reaction of an alkane is: $C_nH_{2n+2} + \frac{3n+1}{2} O_2 	o nCO_2 + (n+1) H_2O$The reaction of $CO_2$ with $NaOH$ to form $Na_2CO_3$ is: $CO_2 + 2NaOH 	o Na_2CO_3 + H_2O$Given $300 \ mL$ of $2 \ N \ NaOH$ solution. Since the $n$-factor for $NaOH$ is $1$,$Normality = Molarity$. Thus,moles of $NaOH = 0.3 \ L 	imes 2 \ M = 0.6 \ mol$.From the stoichiometry of the second reaction,$2 \ mol$ of $NaOH$ react with $1 \ mol$ of $CO_2$. Therefore,moles of $CO_2 = \frac{0.6}{2} = 0.3 \ mol$.From the combustion reaction,$1 \ mol$ of $C_nH_{2n+2}$ produces $n \ mol$ of $CO_2$. Thus,moles of alkane = $\frac{0.3}{n}$.Given the mass of the alkane is $4.3 \ g$ and its molar mass is $(14n + 2) \ g/mol$,we have: $\frac{4.3}{14n + 2} = \frac{0.3}{n}$.Solving for $n$: $4.3n = 0.3(14n + 2) \implies 4.3n = 4.2n + 0.6 \implies 0.1n = 0.6 \implies n = 6$.Thus,the alkane is $C_6H_{14}$.

$4.3 \ g$ of an alkane is burnt in sufficient oxygen. The $CO_2$ formed reacts completely with $300 \ mL, 2 \ N \ NaOH$ solution producing $Na_2CO_3$. The alkane is:

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