What volume of a solution of hydrochloric aci | vedclass

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(A) $1$. The reaction of metallic sodium with water is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$. $2$. Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02

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(A) $1$. The reaction of metallic sodium with water is: $2Na + 2H_2O ightarrow 2NaOH + H_2$. $2$. Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$. $3$. Since $2 \ mol$ of $Na$ produce $2 \ mol$ of $NaOH$,$0.02 \ mol$ of $Na$ will produce $0.02 \ mol$ of $NaOH$. $4$. The neutralization reaction is: $NaOH + HCl ightarrow NaCl + H_2O$. $5$. Moles of $HCl$ required = Moles of $NaOH = 0.02 \ mol$. $6$. The concentration of $HCl$ solution is $73 \ g/L$. Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$. $7$. Molarity of $HCl = \frac{73 \ g/L}{36.5 \ g/mol} = 2 \ M$. $8$. Using $n = M 	imes V(L)$,we get $0.02 = 2 	imes V$. $9$. $V = 0.01 \ L = 10 \ mL$.

What volume of a solution of hydrochloric acid containing $73 \ g$ of acid per litre would be sufficient for the exact neutralization of sodium hydroxide obtained by allowing $0.46 \ g$ metallic sodium to act upon water? $(Cl = 35.5, Na = 23.0, O = 16, H = 1)$

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