posted by Anna on .
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?
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
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.
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.?