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?

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 = ??

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% ?

To answer the question, we need to calculate the theoretical yield of k3(fe(c204)3)*3h2o and the percent yield.

1) To calculate the theoretical yield of k3(fe(c204)3)*3h2o:

The molar mass of ferrous ammonium sulfate (NH4)2Fe(SO4)2·6H2O (also known as FAS) is 392.138 g/mol.
The molar mass of k3(fe(c204)3)·3h2o is 554.373 g/mol.

To find the theoretical yield, we can set up a proportion using the molar masses:

(4g FAS) / (392.138 g/mol FAS) = (x g k3(fe(c204)3)·3h2o) / (554.373 g/mol k3(fe(c204)3)·3h2o)

Simplifying the equation and solving for x, we get:

x = (4g FAS) * (554.373 g/mol k3(fe(c204)3)·3h2o) / (392.138 g/mol FAS)
= 5.66 g

Therefore, the theoretical yield of k3(fe(c204)3)·3h2o is 5.66 grams.

2) To calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) * 100

Given that the actual yield is 4.25 g and the theoretical yield is 5.66 g, we can substitute the values into the formula:

Percent yield = (4.25 g / 5.66 g) * 100
≈ 75.3%

Therefore, the percent yield of k3(fe(c204)3)·3h2o is approximately 75.3%.