Given the reaction C_{(s)} + H_2O_{(l)} to CO | vedclass

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Question

(A) The balanced chemical equation is: $C_{(s)} + H_2O_{(l)} 	o CO_{(g)} + H_{2(g)}$From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO

Vedclass · Accepted Answer

$179.2$

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(A) The balanced chemical equation is: $C_{(s)} + H_2O_{(l)} 	o CO_{(g)} + H_{2(g)}$From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO$ and $1 \ mol$ of $H_2$,totaling $2 \ mol$ of gas.Atomic mass of $C = 12 \ g/mol$.Moles of $C = \frac{48.0 \ g}{12 \ g/mol} = 4 \ mol$.Since $1 \ mol$ of $C$ produces $2 \ mol$ of gas,$4 \ mol$ of $C$ will produce $4 	imes 2 = 8 \ mol$ of gas.At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.Total volume of gases produced = $8 \ mol 	imes 22.4 \ L/mol = 179.2 \ L$.

Given the reaction $C_{(s)} + H_2O_{(l)} \to CO_{(g)} + H_{2(g)}$,calculate the volume of the gases produced at $STP$ from $48.0 \ g$ of carbon.

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