# chemistry question

In the presence of a suitable catalyst, 30.0 cm3
of a 0.10 mol dm–3 aqueous solution of Na2C2O4 reacts exactly with 40.0 cm3 of a 0.15 mol dm–3 aqueous solution of Ce(SO4)2.
The only products of the reaction are another cerium salt, a different water-soluble salt and carbon dioxide.
What is the cerium ion formed by this reaction?

A Ce+
B Ce2+
C Ce3+
D Ce4+
E Ce5+
Original post by hypez
In the presence of a suitable catalyst, 30.0 cm3
of a 0.10 mol dm–3 aqueous solution of Na2C2O4 reacts exactly with 40.0 cm3 of a 0.15 mol dm–3 aqueous solution of Ce(SO4)2.
The only products of the reaction are another cerium salt, a different water-soluble salt and carbon dioxide.
What is the cerium ion formed by this reaction?

A Ce+
B Ce2+
C Ce3+
D Ce4+
E Ce5+

Firstly try to write an equation for the reaction using the information given, you can find mole ratio of reactants and then try to balance. I got C...
Original post by user342
Firstly try to write an equation for the reaction using the information given, you can find mole ratio of reactants and then try to balance. I got C...

what would be the cerium salt?
I was thinking Ce2(SO4)3.
Original post by user342
I was thinking Ce2(SO4)3.

That seems right, I'm just curious as to how you get there.
I found the mole ratio of Na2C2O4:Ce(SO4)2 is 1:2, not sure how i can use this information.
The soluble salt i believe to be Na2SO4, plus CO2 as another product plus I thought Ce(C2O4). Not sure how to start balancing if I don't know the correct products.
Edit: i sort of get it, you need 4 SO4 in products, one comes from Na2SO4 so must be 3 in the Ce salt.
(edited 8 months ago)