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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+
Reply 1
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...
Reply 2
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?
Reply 3
I was thinking Ce2(SO4)3.
Reply 4
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)

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