What volume in mL of 73 g/L hydrochloric aci | vedclass

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Question

(C) The balanced chemical equation for the reaction of sodium with water is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$ Calculate the moles of $Na$: $n(Na)

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) The balanced chemical equation for the reaction of sodium with water is: $2Na + 2H_2O ightarrow 2NaOH + H_2$ Calculate the moles of $Na$: $n(Na) = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$ From the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$,so $0.02 \ mol$ of $Na$ produces $0.02 \ mol$ of $NaOH$. The neutralization reaction is: $NaOH + HCl ightarrow NaCl + H_2O$ Thus,moles of $HCl$ required = moles of $NaOH = 0.02 \ mol$. The molarity of $HCl$ is: $M = \frac{73 \ g/L}{36.5 \ g/mol} = 2 \ M$ Using $n = M 	imes V(L)$: $0.02 = 2 	imes V(L)$ $V(L) = 0.01 \ L$ $V(mL) = 0.01 	imes 1000 = 10 \ mL$.

What volume in $mL$ of $73 \ g/L$ hydrochloric acid is required to completely neutralize the sodium hydroxide produced by the reaction of $0.46 \ g$ of sodium metal with water?

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