# 25 mL of 0.350 M "NaCl" are added to 45 ml of 0.125 M "CuSO"_4. How many grams of "Cu"("OH")_2 will precipitate?

Jul 20, 2017

Here's what I got.

#### Explanation:

I don't think that you have the reactants right here, so I will assume that you're mixing sodium hydroxide, $\text{NaOH}$, and copper(II) sulfate, ${\text{CuSO}}_{4}$.

Start by writing the balanced chemical equation that describes this double replacement reaction

$2 {\text{NaOH"_ ((aq)) + "CuSO"_ (4(aq)) -> "Cu"("OH")_ (2(s)) darr + "Na"_ 2"SO}}_{4 \left(a q\right)}$

Notice that you need $2$ moles of sodium hydroxide in order to react with $1$ mole of copper(II) sulfate and produce $1$ mole of copper(II) hydroxide.

Use the molarities and volumes of the two solutions to figure out how many moles of each reactant you are mixing together.

25 color(red)(cancel(color(black)("mL solution"))) * "0.350 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.00875 moles NaOH"

45color(red)(cancel(color(black)("mL solution"))) * "0.125 moles CuSO"_4/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.005625 moles CuSO"_4

At this point, your goal is to figure out if you have enough moles of sodium hydroxide to ensure that all the moles of copper(II) sulfate will take part in the reaction.

You know that $0.005625$ moles of copper(II) sulfate will require

0.005625 color(red)(cancel(color(black)("moles CuSO"_4))) * "2 moles NaOH"/(1color(red)(cancel(color(black)("mole CuSO"_4)))) = "0.01125 moles NaOH"

overbrace("0.01125 moles NaOH")^(color(blue)("what you need")) " " > " " overbrace("0.00875 moles NaOH")^(color(blue)("what you have"))

This means that sodium hydroxide will act as a limiting reagent, which implies that it will be completely consumed before all the moles of copper(II) sulfate will get the chance to react.

You can thus say that the reaction will consume $0.00875$ moles of sodium hydroxide and produce

0.00875color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole Cu"("OH")_2)/(2color(red)(cancel(color(black)("moles NaOH")))) = "0.004375 moles Cu"("OH")_2

To convert this to grams, use the molar mass of copper(II) hydroxide

$0.004375 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Cu"("OH")_2))) * "97.561 g"/(1color(red)(cancel(color(black)("mole Cu"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.43 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volumes of the solutions.