Which is the limiting and excess reagent in this reaction?
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.
Answer:
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
hehe. chem16?
Limiting Reagent --> Ba(NO3)2
Excess --> Fe2(SO4)3
^^
Limiting Reagent is Ba(NO3)2.
Excess Reagent is Fe2(SO4)3.
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: