Question

Question #12be1

Answer 1

`"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.