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If 0.275 moles of ferrous ammonium sulfate (Fe(NH4)2 *6H2O) nd an excess of all reagents are used in a synthesis of K3[Fe(C2O4)3]*3H2O, how many grams of the product will be obtained if the reaction gives a 100% yield. Show your work for full credit.
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In the molecular formulas of both there is one molecule of Fe hence the number of moles of the compoung will also be 0.275. Molecular weight: 3*39+56*3+(12*2+16*4)*9+3*(2+16). weight= number of moles *Molecular weight. Calculate it yourself.
I did use the molecular weight and calculated it but wanted to make sure it was correct. Thanks for the response though.
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