# Question e3623

Aug 10, 2015

The answer is b) 6 g

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

$N a O {H}_{\left(a q\right)} + H C {l}_{\left(a q\right)} \to N a C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $1 : 1$ mole ratio between sodium hydroxide and hydrochloic acid; this means that a complete neutralization would require equal numbers of moles of sodium hydroxide and hydrochloric acid.

Now, you know that the hydrochloric acid solution has a normality of 0.1 N. Normality is simply a measure of reactivity, meaning that it is calculated by taking into account how a substance behaves in a particular reaction.

Hydrochloric acid dissociates in aqueous solution to rpoduce

$H C {l}_{\left(a q\right)} \to {H}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

The net ionic equation for your reaction will be

$O {H}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+} \to {H}_{2} {O}_{\left(l\right)}$

In your case, a 0.1 N solution means that the hydrochloric acid solution provides 0.1 moles of protons, ${H}^{+}$, per liter to the reaction.

Since 1 mole of $\text{HCl}$ produces 1 mole of ${H}^{+}$, the molarity of the solution will be equal to $\text{0.1 mol/L}$.

The volume of the solution will be

1500color(red)(cancel(color(black)("cm"^3))) * ("1 dm"""^3)/(1000color(red)(cancel(color(black)("cm"^3)))) = "1.5 dm"""^3

This means that you can now calculate how many moles of hydrochloric acid took part in the reaction (remember that $\text{1 dm"^3 = "1 L}$)

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{HCl" = 0.1 "moles"/color(red)(cancel(color(black)("L"))) * 1.5color(red)(cancel(color(black)("L"))) = "0.15 moles HCl}}$

The aforementioned mole ratio tells you that the number of moles of sodium hydroxide needed to neutralize this many moles of hydrochloric acid is

${n}_{N a O H} = {n}_{H C l} = \text{0.15 moles}$

To get the mass of sodium hydroxide needed, use its molar mass

0.15color(red)(cancel(color(black)("moles"))) * "40.0 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("6 g")#