Question

Question #e3623

Answer 1

The answer is b) 6 g

Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

`NaOH_((aq)) + HCl_((aq)) -> NaCl_((aq)) + H_2O_((l))`

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

`HCl_((aq)) -> H_((aq))^(+) + Cl_((aq))^(-)`

The net ionic equation for your reaction will be

`OH_((aq))^(-) + H_((aq))^(+) -> H_2O_((l))`

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 `"HCl"` produces 1 mole of `H^(+)`, the molarity of the solution will be equal to `"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 `"1 dm"^3 = "1 L"`)

`C = n/V implies n = C * V`

`n_"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_(NaOH) = n_(HCl) = "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")`