When 1 L of CO_2 is heated with graphite,the | vedclass

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(B) The chemical reaction is: $CO_{2(g)} + C_{(s)} \xrightarrow{\Delta} 2CO_{(g)}$Let the volume of $CO_2$ reacted be $x \ L$.Initially: $1 \ L$ of $C

Vedclass · Accepted Answer

$0.0714$

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(B) The chemical reaction is: $CO_{2(g)} + C_{(s)} \xrightarrow{\Delta} 2CO_{(g)}$Let the volume of $CO_2$ reacted be $x \ L$.Initially: $1 \ L$ of $CO_2$ is taken.After reaction: Volume of $CO_2$ remaining $= (1-x) \ L$ and volume of $CO$ produced $= 2x \ L$.Total volume of gases $= (1-x) + 2x = 1 + x = 1.8 \ L$.Therefore,$x = 0.8 \ L$.Volume of $CO$ produced $= 2 	imes 0.8 = 1.6 \ L$.At $STP$,$1 \ mole$ of gas occupies $22.4 \ L$.Number of moles of $CO = \frac{1.6 \ L}{22.4 \ L/mol} \approx 0.0714 \ mol$.

When $1 \ L$ of $CO_2$ is heated with graphite,the volume of the gases collected is $1.8 \ L$. What will be the number of moles of $CO$ produced at $STP$?

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