The volume of 0.1 , M H_2SO_4 that is needed | vedclass

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Question

(C) The balanced chemical equation for the reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. According to the stoichiometry,$1 \, 	ext{mol

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The balanced chemical equation for the reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. According to the stoichiometry,$1 \, 	ext{mole}$ of $H_2SO_4$ reacts with $2 \, 	ext{moles}$ of $NaOH$. Using the formula $M_1 V_1 n_2 = M_2 V_2 n_1$,where $n_1$ and $n_2$ are the stoichiometric coefficients: $0.1 	imes V_1 	imes 2 = 0.2 	imes 40 	imes 1$. $0.2 	imes V_1 = 8$. $V_1 = \frac{8}{0.2} = 40 \, mL$. Thus,the correct option is $(c)$.

The volume of $0.1 \, M$ $H_2SO_4$ that is needed to completely neutralize $40 \, mL$ of $0.2 \, M$ $NaOH$ is $..... \, mL$.

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