What will be the volume of O_2 gas at STP for | vedclass

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(D) The balanced chemical equation for the combustion of acetylene $(C_2H_2)$ is: $C_2H_2 + \frac{5}{2} O_2 \rightarrow 2 CO_2 + H_2O$ First,calculate

Vedclass · Accepted Answer

$16.8$

Vedclass · Answer

(D) The balanced chemical equation for the combustion of acetylene $(C_2H_2)$ is: $C_2H_2 + \frac{5}{2} O_2 ightarrow 2 CO_2 + H_2O$ First,calculate the number of moles of acetylene: $n(C_2H_2) = \frac{	ext{mass}}{	ext{molar mass}} = \frac{7.8 \ g}{26 \ g/mol} = 0.3 \ mol$ According to the stoichiometry,$1 \ mol$ of $C_2H_2$ requires $\frac{5}{2} \ mol$ of $O_2$. Therefore,$0.3 \ mol$ of $C_2H_2$ requires $0.3 	imes 2.5 = 0.75 \ mol$ of $O_2$. At $STP$,the volume of $1 \ mol$ of gas is $22.4 \ L$. $V(O_2) = 0.75 \ mol 	imes 22.4 \ L/mol = 16.8 \ L$

What will be the volume of $O_2$ gas at $STP$ for complete combustion of $7.8 \ g$ acetylene? ............ $L$

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