Complete combustion of 3 g of ethane gives x | vedclass

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(D) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is:$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$Molar mass of etha

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(D) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is:$2C_2H_6(g) + 7O_2(g) ightarrow 4CO_2(g) + 6H_2O(l)$Molar mass of ethane $(C_2H_6)$ = $(2 	imes 12.0) + (6 	imes 1.0) = 30 \ g/mol$.Moles of ethane = $\frac{3 \ g}{30 \ g/mol} = 0.1 \ mol$.From the stoichiometry of the reaction,$2 \ mol$ of $C_2H_6$ produces $6 \ mol$ of $H_2O$.Therefore,$1 \ mol$ of $C_2H_6$ produces $3 \ mol$ of $H_2O$.So,$0.1 \ mol$ of $C_2H_6$ produces $0.1 	imes 3 = 0.3 \ mol$ of $H_2O$.Number of molecules of water = $	ext{moles} 	imes N_A = 0.3 	imes 6.023 	imes 10^{23} = 1.8069 	imes 10^{23} = 18.069 	imes 10^{22}$.Rounding to the nearest integer,the value of $x$ is $18$.

Complete combustion of $3 \ g$ of ethane gives $x \times 10^{22}$ molecules of water. The value of $x$ is ......... (Round off to the Nearest Integer). [Use: $N_{A} = 6.023 \times 10^{23} ;$ Atomic masses in $u$ $C = 12.0, O = 16.0, H = 1.0$]

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