Answer to Question #199624 in General Chemistry for Allison

Question

If 86.7 g of Fe react in an excess of water, how many grams of Fe₃O₄ are produced?

 

3 Fe   +   4 H₂O  -------->   Fe₃O₄     +     4 H₂

Answer 1

MM(Fe) = 55.84 g/mol

$n(Fe) = \frac{86.7}{55.84}=1.552 \;mol$

According to the reaction:

n(Fe₃O₄) $= \frac{1}{3}n(Fe) = 0.517 \;mol$

MM(Fe₃O₄)=231.55 g/mol

m(Fe₃O₄) $= 0.517 \times 231.55 = 359.5 \;g$

Answer: 359.5 g