# Question 1c600

Feb 6, 2017

Here's what I got.

#### Explanation:

All you really have to do here is to backtrack from the mass of copper(II) oxide produced by the second reaction to the mass of copper(II) sulfate that reacted in the first reaction by using mole ratios.

You know that

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

and

${\text{Cu"("OH")_ (2(s)) -> "CuO"_ ((s)) + "H"_ 2"O}}_{\left(l\right)}$

Notice that copper(II) hydroxide and copper(II) oxide have a $1 : 1$ mole ratio in the second reaction.

For a 100% yield reaction, this implies that the number of moles of copper(II) oxide produced by the reaction will be equal to the number of moles of copper(II) hydroxide that underwent decomposition.

Use the molar mass of copper(II) oxide to find the number of moles produced by the second reaction

0.1080 color(red)(cancel(color(black)("g"))) * "1 mole CuO"/(79.545color(red)(cancel(color(black)("g")))) = "0.0013577 moles CuO"

This means that the reaction consumed $0.0013577$ moles of copper(II) hydroxide.

Now, look at the first reaction. Copper(II) hydroxide and copper(II) sulfate have a $1 : 1$ mole ratio, so the same principle applies -- the number of moles of the former produced by the reaction will be equal to the number of moles of the latter consumed by the reaction.

Therefore, the first reaction consumed $0.0013577$ moles of copper(II) sulfate.

Notice that the sodium hydroxide is in excess because the solution contains

15.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "6.0 moles NaOH"/(1color(red)(cancel(color(black)("L solution")))) = "0.090 moles NaOH"

In order for all the moles of copper(II) sulfate to react, you only need

0.0013577 color(red)(cancel(color(black)("moles Cu"("SO")_4))) * "2 moles NaOH"/(1color(red)(cancel(color(black)("mole Cu"("SO")_4)))) = "0.0027154 moles NaOH"#

This confirms that the first reaction will indeed consume $0.0013577$ moles of copper(II) sulfate. Use the compound's molar mass to convert the number of moles to grams

$0.0013577 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles Cu"("SO")_4))) * "159.609 g"/(1color(red)(cancel(color(black)("mole Cu"("SO")_4)))) = color(darkgreen)(ul(color(black)("0.2167 g CuSO}}_{4}}}}$

I'll leave the answer rounded to four sig figs, but keep in mind that you only have two sig figs for the molarity and volume of the sodium hydroxide solution.

The actual number of moles of sodium hydroxide is not used in the calculations, so I'll not round the answer to two sig figs.