A piece of iron was placed in a silver nitrate solution. After the reaction iron's mass increased by 3.52g. How many moles of #"Fe"("NO"_3)_2# were produced?

1 Answer
Jul 24, 2016

Answer:

#"0.0220 moles"#

Explanation:

You're dealing with a redox reaction in which iron, #"Fe"#, reduces the silver cations, #"Ag"^(+)#, to silver metal, #"Ag"#, while being oxidized to iron(II) cations, #"Fe"^(2+)#.

The net ionic equation that describes this reaction looks like this

#"Fe"_ ((s)) + 2"Ag"_ ((aq))^(+) -> "Fe"_ ((aq))^(2+) + color(red)(2)"Ag"_ ((s)) darr#

The silver metal produced by the reaction will be deposited on the piece of iron. As a result, its mass will increase.

Now, here's where the tricky part comes in. Keep in mind that some of the iron will be converted to iron(II) cations. This means that if you disregard the deposited silver, the piece of iron will actually have a smaller mass once the reaction is completed.

Let's take #m_"Fe"# #"g"# to be the initial mass of the piece of iron. Let's assume that #x# represents the number of moles of iron(II) cations produced by the reaction. Iron has a molar mass of #"55.845 g mol"^(-1)#.

This means that the mass of iron that was converted to iron(II) cations was

#x color(red)(cancel(color(black)("moles Fe"^(2+)))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe"^(2+))))) = (55.845 * x)" g"#

The mass of the piece of iron after the reaction is completed, once again disregarding the mass of silver deposited, will be

#m_"Fe" " g" - (x * 55.845)" g" = (m_"Fe" - x * 55.845)" g"#

Now, notice that for every mole of iron(II) cations produced by the reaction, you also get #color(red)(2)# moles of silver metal. It follows that for #x# moles of iron(II) cations, the reaction will produce #(color(red)(2)x)# moles of silver metal.

Use the molar mass of silver to find the mass of silver metal produced by the reaction

#(color(red)(2)x) color(red)(cancel(color(black)("moles Ag"))) * "107.87 g"/(1color(red)(cancel(color(black)("mole Ag")))) = (215.74 * x)" g Ag"#

This mass is added to the mass of iron to get the total mass of the piece of iron, i.e. iron + deposited silver

#overbrace((m_"Fe" - 55.845 * x) " g")^(color(blue)("remaining mass of iron")) + overbrace((215.74 * x)" g")^(color(purple)("mass of silver deposited")) = overbrace((m_"Fe" + 3.52)" g")^(color(darkgreen)("mass of iron + silver"))#

This is equivalent to

#color(red)(cancel(color(black)(m_"Fe"))) - 55.845 * x + 215.74 * x = color(red)(cancel(color(black)(m_"Fe"))) + 3.52#

#159.895 * x = 3.52 implies x = 3.52/159.895 = 0.0220#

Since #x# represents the number of moles of iron(II) cations, which keep in mind is equal to the number of moles of iron(II) nitrate, #"Fe"("NO"_3)_2#, produced by the reaction

#"Fe"("NO"_ 3)_ (2(aq)) -> "Fe"_ ((aq))^(2+) + 2"NO"_ (3(aq))#

you will have

#"no. of moles of Fe"^(2+) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0220 moles")color(white)(a/a)|)))#

The answer is rounded to three sig figs.