If .275 moles of ferrous ammonium sulfate (Fe(NH4)2(SO4)2. 6H2O) 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 100% yield?

Question

Can someone help me solve this I found the molar mass of (Fe(NH4)2(SO4)2. 6H2O) = 392.1g/mol and K3[Fe(C2O4)3].3H2O=491.1 g/mol
Thank You

Answer 1

since they each have 1 Fe, 1 mole of the Fe(NH4)2(SO4)2· 6H2O makes 1 mole of the K3[Fe(C2O4)3]·3H2O

so, since you start with .275 moles of (Fe(NH4)2(SO4)2. 6H2O) you end up with .275 moles of K3[Fe(C2O4)3].3H2O.

Now change that to grams and you're done.

do a web search for your question, and you'll find the problem answered.

Answer 2

Question...

So then is it unnecessary to calculate the molar mass of the reagent.?
Is only the molar mass of the product relevant to this question.?