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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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RJ5876 RJ5876
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11 years ago
I'd be grateful to whoever can answer all of this or any of these questions.

1.48 g of ferric sulfate reacts with 60 mL of .1002 M barium nitrate.
Which is the limiting reagent? The excess reagent?
How many moles of the excess reagent will remain?
What weight of barium sulfate will be produced from the reaction?


Here is what I think is the balanced chemical equation:

Fe2(SO4)3 + 3Ba(NO3)2 ---> 3BaSO4 + 2Fe(NO3)3


I'm desperate. Please help me. Thank you.
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wrote...
#1
11 years ago
hehe. chem16?
Limiting Reagent Rightwards Arrow Ba(NO3)2
Excess Rightwards Arrow Fe2(SO4)3

Upwards Arrow
wrote...
#2
11 years ago
Limiting Reagent is Ba(NO3)2.
Excess Reagent is Fe2(SO4)3.
wrote...
#3
11 years ago
Molecular weight Fe2(SO4)3 is : 399.7 g/mol
1.48 g / 399.7 = 0.00370 moles
60 x 0.1002 /1000 = 0.00601 moles Ba(NO3)2
the ratio between Fe2(SO4) and Ba(NO3)2 is 1 : 3 so Ba(NO3)2 is the limiting reactant   and Fe2(SO4)3 is the  reactant  in excess
1 : 3 = x :  0.00601
x = 0.00200 moles Fe2(SO4)3 needed for the reaction
0.00370 - 0.00200 = 0.00170 moles Fe2(SO4)3 in excess
the ratio between Fe2(SO4)3 and BaSO4 is 1 : 3 so we get 0.00370 x 3 = 0.0111 moles BaSO4
Molecular weight = 233 g/mol
0.0111 mol x 233 g/mol = 2.59 g
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