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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
Thank You

  • Science -

    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.

  • Science -


    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.?

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