Question #12be1

1 Answer
Jun 12, 2016

Answer:

#"87.9 g AgNO"_3#

Explanation:

Start by writing the balanced chemical equation that describes this double replacement reaction

#color(red)(2)"AgNO"_ (3(aq)) + "FeCl"_ (2(aq)) -> 2"AgCl"_ ((s)) darr + "Fe"("NO"_ 3)_ (2(aq))#

Here you have #color(red)(2)# moles of silver nitrate, #"AgNO"_3#, reacting with #1# mole of iron(II) chloride, #"FeCl"_2#, to produce #2# moles of silver chloride and #1# mole of aqueous iron(II) nitrate.

You know that the two reactants are used up in a #color(red)(2):1# mole ratio, which means that you can use their molar masses to convert this into a gram ratio.

You will have

#color(red)(2)color(red)(cancel(color(black)("moles AgNO"_3))) * "169.87 g"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "339.74 g"#

and

#1color(red)(cancel(color(black)("mole FeCl"_2))) * "126.75 g"/(1color(red)(cancel(color(black)("mole FeCl"_2)))) = "126.75 g"#

Therefore, the #color(red)(2):1# mole ratio is equivalent to a #339.74 : 126.75# gram ratio.

Since you know that the reaction must consume #"32.8 g"# of iron(II) chloride, you can say that you need

#32.8 color(red)(cancel(color(black)("g FeCl"_2))) * "339.74 g AgNO"_3/(126.75color(red)(cancel(color(black)("g FeCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("87.9 g AgNO"_3)color(white)(a/a)|)))#

The answer is rounded to three sig figs.