Question

Question #f2bc2

Answer 1

`5 * 10^(-3)"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

`"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, `"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 `"0.2 M"`, or `"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

`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 `"1 L"` of solution contains `0.2` 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.