To neutralize 25 , mL of 0.25 , M Na_2CO_3 so | vedclass

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(B) The balanced chemical equation for the reaction is: $Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$. According to the stoichiometry,$1 \, 	ext{

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$25$

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(B) The balanced chemical equation for the reaction is: $Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$. According to the stoichiometry,$1 \, 	ext{mole}$ of $Na_2CO_3$ reacts with $2 \, 	ext{moles}$ of $HCl$. Using the formula $n_1M_1V_1 = n_2M_2V_2$ (where $n$ is the n-factor or stoichiometric coefficient ratio): $n_1 = 1$ (for $Na_2CO_3$),$M_1 = 0.25 \, M$,$V_1 = 25 \, mL$. $n_2 = 2$ (for $HCl$),$M_2 = 0.5 \, M$,$V_2 = ?$. $1 	imes 0.25 	imes 25 = 2 	imes 0.5 	imes V_2$. $6.25 = 1 	imes V_2$. $V_2 = 25 \, mL$.

To neutralize $25 \, mL$ of $0.25 \, M$ $Na_2CO_3$ solution,how much volume of $0.5 \, M$ $HCl$ is required? (Note: The reaction is $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$)

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