The volume of oxygen at STP required to burn | vedclass

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Question

(D) The combustion reaction of carbon is: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$ From the stoichiometry,$1 \,mole$ of carbon $(12 \,g)$ reacts

Vedclass · Accepted Answer

$4.48$

Vedclass · Answer

(D) The combustion reaction of carbon is: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$ From the stoichiometry,$1 \,mole$ of carbon $(12 \,g)$ reacts with $1 \,mole$ of oxygen ($22.4 \,L$ at $STP$). Therefore,$2.4 \,g$ of carbon reacts with: $	ext{Moles of } C = \frac{2.4 \,g}{12 \,g/mol} = 0.2 \,mol$ Since $1 \,mol$ of $C$ requires $1 \,mol$ of $O_2$,$0.2 \,mol$ of $C$ requires $0.2 \,mol$ of $O_2$. Volume of $O_2$ at $STP = 0.2 \,mol 	imes 22.4 \,L/mol = 4.48 \,L$.

The volume of oxygen at $STP$ required to burn $2.4 \,g$ of carbon completely is $.... \,L$

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