The number of molecules of CO_2 liberated by | vedclass

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(C) The combustion reaction of graphite is given by:$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$From the stoichiometry of the reaction,$1 	ext{ mol

Vedclass · Accepted Answer

$6.02 	imes 10^{22}$

Vedclass · Answer

(C) The combustion reaction of graphite is given by:$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$From the stoichiometry of the reaction,$1 	ext{ mole}$ of $C$ produces $1 	ext{ mole}$ of $CO_2$.Since $1 	ext{ mole}$ of any substance contains $6.022 	imes 10^{23}$ molecules:$1 	ext{ mole of } C 	ext{ produces } 6.022 	imes 10^{23} 	ext{ molecules of } CO_2$.Therefore,$0.1 	ext{ mole}$ of graphite will produce:$0.1 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{22} 	ext{ molecules of } CO_2$.

The number of molecules of $CO_2$ liberated by the complete combustion of $0.1 \text{ mol}$ of graphite in air is:

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