Question

I NEED HELP ASAP!!! Will give a medal!

76.59 g of chromium metal reacts with an excess of zinc sulfate to produce 144.51 g of zinc metal.

Calculate the molar ratio of these two metals.

Answer 1

@Cuanchi

Answer 2

The molar ratio is shown in the balanced equation. So all you need to do is write the names into formulas, write the equation and balance it. The ratio of the coefficients of the two metals will be the molar ratio they ask for.

Answer 3

So
\[Cr+ZnSO _{4}\rightarrow Zn\]

Answer 4

The final product doesn't have Cr or SO4 in it though... so how do I balance it?

Answer 5

It's sort of implied in the question that Cr and \(\sf SO^{2-}_4\) will for a compound after Cr donates electrons to \(\sf Zn^{2+}\), it's a redox reaction. It can also be classified as a single-displacement reaction.

Sorry, lets backtrack. I didn't realize they wanted you to find the oxidation state of Cr, based on how many electrons it loses.

So it's basic stoichiometry. Find the moles of Cr metal and Zn metal, then use their moles (expressed as simplest form) as the ratio they ask for.

Answer 6

Btw, \(\sf moles=\dfrac{mass}{molar~mass}\)

Answer 7

Aren't both of their moles just 1 because there isn't a coefficient?

Answer 8

The problem is that we dont know the formula of \(\sf Cr_x(SO_4)_y\) based on how many electrons Cr loses.

It could lose 1 and be \(Cr_2SO_4\), or it could lose 2 and be \(CrSO_4\) or even lose 3 and be \(Cr_2(SO_4)_3\).

Answer 9

Transition metals can vary on their oxidation states and without a reduction potential table, you wont know how many electrons it loses.

Answer 10

So how exactly am I supposed to solve this then? It doesn't specify how many electrons Cr loses. I am supposed to compare it in the next question though. It asks if the mole ratio supports \[Cr+ ZnSO _{4}\rightarrow Zn+ CrSO _{4}\] or \[2Cr+ 3ZnSO _{4}\rightarrow 3Zn+ Cr _{2}(SO _{4})_{3}\] does this information help me solve it at all?

Answer 11

yes it does.

You have to do some stoichiometry, basically find the theoretical yield (were assuming the reaction went to 100%) with the starting material (76 g Cr) - try both sets of coefficients to see which one is actually right - and see which one gives you the mass of Zn metal you got.

Answer 12

Convert mass of Cr to moles, and use the coefficients of Cr and Zn to find the moles of Zn produced.

\(\sf \dfrac{moles ~of ~Cr}{Cr's~coefficient}=\dfrac{moles~of~Zn}{Zn's~coefficient}\)

\(\sf moles~of~Zn=\dfrac{(moles ~of ~Cr)*(Zn's~coefficient)}{Cr's~coefficient}\)

Answer 13

It's trial and error with the sets of coefficients from the two reactions.

Answer 14

So I use trial and error for Cr's coefficients and Zn's coefficients? Do I just use 1 mole for the moles of Cr and Zn then?

Answer 15

Or do I have to go from grams to moles to find the amount of moles?

Answer 16

yeah, trial and error to see which set of coefficients matches your data.

Use the moles of Cr that 76.59 g are equal to.

Also, I would convert the mass of Zn you got to moles for comparison - but your not using this in the equation above since it's your unknown.

Answer 17

Okay I think I get it now! Thanks! I'll give it a try and if I still can't get it I will let you know. :)

Answer 18

Okay awesome! no problem :)

Answer 19

How do I find the coefficient on \[ZnSO _{4}\]

Answer 20

once you find the coefficients for Cr and Zn, just balance the reaction.

Answer 21

I found the coefficient for Zn and it didn't match either of the two equations I mentioned earlier. I got 1.473 moles of Cr and 2.2103 moles of Zn

Answer 22

so you can round, to 1.5 and 2. try it with those. sorry i have to go. i'll be back in an hour or so and will help if someone else hasn't by then, sorry!

Answer 23

The options for the coefficient on Zn is either 1 or 3. 2 isn't one of the options so I don't know where I wen't wrong. Thanks again I appreciate the time you put in to help me.

Answer 24

So instead of trying each individual set of coefficients, i'm just going to use the moles used and obtained - this is actually the first method i mentioned (in my second post) that you should do, it saves you time.

\(\sf \dfrac{moles~of~Cr}{Cr's~coefficient}=\dfrac{moles~of~Zn}{Zn's~coefficient} \rightarrow \dfrac{0.679~moles}{Cr's~coefficient}=\dfrac{0.452}{1}\)

I set Zn's coefficient to 1, to be able to solve for Cr's coefficient. This is valid because their coefficients are relative, you could've chosen any number but 1 is easy to work with.

Cr's coefficient = 1.5

Now going back to the equation,

\(\sf 1.5~Cr+ ZnSO _{4}\rightarrow Zn+ Cr_{1.5}SO _{4}\)

We need whole numbers in the equation, so we multiply by everything by 2 to normalize (statistically speaking).

\(\sf 3~Cr+ 2~ZnSO _{4}\rightarrow 2~Zn+ Cr_3(SO _{4})_2\)

So now, we know that Cr lost 3 electrons. The molar ratio is Cr:Zn, 3:2.

I know you said 2 isn't an option for Zn, but that's what the data says.

Answer 25

Let me know if you have doubts.