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?
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...
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...
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.