Asked by eddy
can someone please help me with this question:
mass of ferrous ammonium sulfate = 4g
mass of k3(fe(c204)3)*3h2o = 4.25 g
1) using the above mass of ferrous ammonium sulfate calculate the theoretical yield of k3(fe(c204)3)*3h2o?
2) the percent yeild of it?
mass of ferrous ammonium sulfate = 4g
mass of k3(fe(c204)3)*3h2o = 4.25 g
1) using the above mass of ferrous ammonium sulfate calculate the theoretical yield of k3(fe(c204)3)*3h2o?
2) the percent yeild of it?
Answers
Answered by
DrBob222
If this is all of the information you have (I saw a similar problem earlier that listed three equations to go along with the other material), here is a shortcut.
Let's call Fe(NH4)2(SO4)2 "molar mass 1 = M1) and K3[Fe(C2O4)3.3H2O "molar mass 2 = M2).
Then 4.0 x (M2/M1) = g K3[Fe(C2O4)3.3H2O = theoretical yield.
% yield = (4.25/theoretical yield)*100 = ??
Let's call Fe(NH4)2(SO4)2 "molar mass 1 = M1) and K3[Fe(C2O4)3.3H2O "molar mass 2 = M2).
Then 4.0 x (M2/M1) = g K3[Fe(C2O4)3.3H2O = theoretical yield.
% yield = (4.25/theoretical yield)*100 = ??
Answered by
popeye
what mass of barium sulfate can be recovered from mixing 48.75 mL of 0.1872 barium cloride with 27.5mL of 0.3500M sulfuric acid if the % yield is 81.7% ?
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