# Question #f2bc2

##### 1 Answer

#### Answer:

#### 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

#"NaOH"_ ((aq)) + 2"HCl"_ ((aq)) -> "H"_ 2"O"_ ((l)) + 2"NaCl"_((aq))#

Now, the problem wants you to use the **molarity** and volume of the sodium hydroxide solution to determine how many *moles* of sodium hydroxide,

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 **moles** of solute per liter of solution.

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(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 **moles** of sodium hydroxide, it follows that this volume will contain

#2.50 * 10^(-2) color(red)(cancel(color(black)("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"color(white)(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*.