Question

100 mL of 0.10 molL-1 CuSO4 solution is mixed with 50 mL of 0.40 molL-1 NaOH solution and reacts according to the equation below

CuSO4 + 2NaOH -> Cu(OH)2 + Na2SO4

Determine the mass (in g) of Cu(OH)2 produced.

Answer 1

How is it coming so far?

Answer 2

Stuck on this question

Answer 3

Not even sure where to start

Answer 4

First find the moles of each reactant using the volume and concentrations.

Answer 5

Watch your units though

Answer 6

n(CuSO4)=0.10molL-1x0.1L=0.01mol
n(NaOH)=0.40molL-1x0.05L=0.02mol

is that right? what do I do next?

Answer 7

You got it! You now have moles of reactants. You can find your limiting reactant!

Answer 8

0.01molCuSO4x2molNaOH/1molCuSO4=0.02mol NaOH needed
0.02molNaohx1molCuSO4/2molNaOH=0.01mol CuSO4 needed
?

Answer 9

So, it looks like we are at molar ratios or that both are limiting.

Answer 10

Alright, so from the information I have, can I calculate the mass of Cu(OH)2 produced?

Answer 11

Use either amount of moles to calculate the moles of Cu(OH)2 using the molar ratio.

Answer 12

Then the molar mass to find the mass.

Answer 13

Okay, so as the molar ratio of CuSO4:Cu(OH)2 is 1:1, is the number of moles of Cu(OH)2 0.01?

Answer 14

Right!

Answer 15

And the mass?

Answer 16

Therefore, the mass of Cu(OH)2 = 0.01molx97.56gmol-1=0.9756 or 0.976g

Answer 17

Good job :)

Answer 18

Thank you! I actually understand it now I think :)

Answer 19

Awesome! That is the goal. :)