Question

If .275 moles of ferrous ammonium sulfate and an excess of all other reagents are used in a synthesis of
K3[Fe(C2O4)3]*3H2O, how many grams of product will be obtained if the reaction gives a 53% yield?

Can you plz explain to me in steps

Answer 1

I would figure out how many moles of iron is in .275 moles of ferrous ammonium sulfate.

Then, that same number of mole times .53 will be the moles of iron in the product.

Moles product= moles iron in the product.
Grams product=molesproduct*MolMass product.

There are harder ways to find the result.