# Question f2bc2

Apr 23, 2016

$5 \cdot {10}^{- 3} \text{moles NaOH}$

#### Explanation:

Even thought you don't need a balanced chemical equation to answer this question, it's always a good idea to make sure that you're only working with balanced chemical equations.

In this case, you would have

${\text{NaOH"_ ((aq)) + 2"HCl"_ ((aq)) -> "H"_ 2"O"_ ((l)) + 2"NaCl}}_{\left(a q\right)}$

Now, the problem wants you to use the molarity and volume of the sodium hydroxide solution to determine how many moles of sodium hydroxide, $\text{NaOH}$, it contains.

The thing to remember about molarity is that it can be used as conversion factor to help you go from moles to liters of solution, and vice versa.

A solution's molarity tells you how many moles of solute you get per liter of solution. A molarity of $\text{0.2 M}$, or ${\text{0.2 mol L}}^{- 1}$, will contain $0.2$ moles of solute per liter of solution.

Since you're working with a volume expressed in milliliters, use the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 L" = 10^3"mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to convert it to liters. You will have

25.0 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 2.50 * 10^(-2)"L"#

So, if $\text{1 L}$ of solution contains $0.2$ moles of sodium hydroxide, it follows that this volume will contain

$2.50 \cdot {10}^{- 2} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L solution"))) * overbrace("0.2 moles NaOH"/(1color(red)(cancel(color(black)("L solution")))))^(color(purple)("= 0.2 M")) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-3)"moles NaOH} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to one sig fig, the number of sig figs you have for the molarity of the solution, and written in scientific notation.