Question #86038
How many milliliters of an aqueous solution of 0.126 M iron(II) nitrate is needed to obtain 16.3 grams of the salt?

________ mL
1
Expert's answer
2019-03-15T02:34:04-0400

n(Fe(NO3)2)=16.3g179.85g/mol=0.09mol\dfrac{16.3g}{179.85g/mol} = 0.09 mol

V=1000mL∗0.09mol0.126M=714.28mL\dfrac{1000mL*0.09 mol}{0.126M} = 714.28 mL


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