When 100 , ml of 0.1 , M H_2SO_4 reacts with | vedclass

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(C) The balanced chemical equation for the reaction is:
$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$
From the stoichiometry of the reacti

Vedclass · Accepted Answer

$0.224$

Vedclass · Answer

(C) The balanced chemical equation for the reaction is:
$Na_2CO_3 + H_2SO_4 ightarrow Na_2SO_4 + H_2O + CO_2$
From the stoichiometry of the reaction, $1 \, mole$ of $H_2SO_4$ produces $1 \, mole$ of $CO_2$.
First, calculate the number of moles of $H_2SO_4$:
$Moles = Molarity 	imes Volume \, (in \, L) = 0.1 \, M 	imes 0.1 \, L = 0.01 \, moles$.
Since $1 \, mole$ of $H_2SO_4$ produces $1 \, mole$ of $CO_2$, $0.01 \, moles$ of $H_2SO_4$ will produce $0.01 \, moles$ of $CO_2$.
At $STP$, $1 \, mole$ of any gas occupies $22.4 \, L$.
Therefore, the volume of $CO_2$ produced is:
$Volume = 0.01 \, moles 	imes 22.4 \, L/mole = 0.224 \, L$.

When $100 \, ml$ of $0.1 \, M$ $H_2SO_4$ reacts with $Na_2CO_3$, how many liters of $CO_2$ will be produced at $STP$ (in $L$)?

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