ADA Sample Test - G. Chem. Q # 51

[fleur531]

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Does anyone know how to solve this problem??

During a titration it was determined that 30.00
mL of a 0.100 M Ce4+ solution was required to
react completely with 20.00 mL of a 0.150 M
Fe2+ solution. Which one of the following reactions
occurred?

A. Ce4' + 3Fe2+ + H20 -+ 3Fe3+ +
CeO- + 2H+

B. 2Ce4+ + Fe2+ +>> Fe4+ + 2Ce3+

C. Ce4+ + Fe2+ -->> Fe3+ + Ce3+

D. Ce4+ + 2F2+ ->> 2Fe3+ + Ce2+

E* ce4+ + 2Fe2+ >>>> 2 Fe 4 + + Cs2+ + 2e-

The correct answer is indicated as C. I can't seem to figure out why though.

Thanks a bunch in advance!!!

 

[Maximilian]

It gives you the volume and Molarity of each solution used in the reaction. If multiply volume x Molarity, you get the number of moles of each solution. In both cases, Ce4 and Fe2, the number of moles is 0.003. Therefore, they react with a 1:1 mole ratio.

Answer C is the only choice with a 1:1 mole ratio.

 

[MattH25]

Use the info to convert them both to moles reacted.

30.00mL Ce (0.100mol/1000mL) = 0.003 moles reacted

20.00mL Fe (0.150mol/1000mL) = 0.003 moles reacted

It's a 1:1 ratio so "C" is the only correct answer

You can run a quick calculation just to double check your answer...
30.00mL Ce (0.100mol Ce/1000mL)(1mol Fe/1mol Ce)(1000mL Fe/0.150mol Fe) = 20mL Fe

 

[fleur531]

makes total sense now... thank you both so much!!!