If 0.275 moles of ferrous ammonium sulfate and an excess of allother reagents ar

Question

If 0.275 moles of ferrous ammonium sulfate and an excess of allother reagents are used in a synthesis ofK3[Fe(C2O4)3]*3H2O, how many grams of product will be obtained ifthe reaction gives a 53% yield?

Explanation / Answer

This is how its done, use stoichiometry setup this up Fe(II)[NH4SO4]2 + EXCESS Reagents-> K3[Fe(C2O4)3]*3H2O It is safe to assume the mole ratio is 1-1. This is eveneasier to calculate .275 mol Fe(II)[NH4SO4]2 x ( 1 mol K3[Fe(C2O4)3]*3H2O / 1 molFe(II)[NH4SO4]2 ) x ( 491.1 gK3[Fe(C2O4)3]*3H2O / 1 molK3[Fe(C2O4)3]*3H2O ) = 135.05 gK3[Fe(C2O4)3]*3H2O and because of 53% yield use the following concept - >     x / 135.05 gK3[Fe(C2O4)3]*3H2O = .53 x = 71.58 gK3[Fe(C2O4)3]*3H2O is obtained Hope this answers ur question. Please remember to rate myanswer. Ty

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