If 5.04 grams of iron (II) nitrate reacts with an excess of #H_2S#, what mass of iron(II) sulfide will be produced in the reaction #Fe(NO_3)_2 + H_2S -> FeS + H(NO_3)#?

1 Answer
Aug 8, 2016

Answer:

Mass of ferrous nitrate #=# #(5.04*g)/(179.86*g*mol^-1)xx87.91*g*mol^-1# #=# #??g#

Explanation:

#Fe(NO_3)_2(aq) + H_2S(aq) rarr FeS(s)darr + 2HNO_3(aq)#

The stoichiometrically balanced equation tells you explicitly that 1 mol of sulfide will result from every mole of ferrous nitrate.

#"Moles of ferrous nitrate"# #=# #(5.04*g)/(179.86*g*mol^-1)# #=# #??#

#"Mass of ferrous sulfide"# #=# #(5.04*g)/(179.86*g*mol^-1)xx87.91*g*mol^-1# #~=# #2.0*g#

The sulfide will be obtained as a sticky, black mess that is exceptionally difficult to handle and dry.