Question

When 13.55 grams of Magnesium metal reacts with 80.50 grams of iron (II) chloride, what is the limiting reactant?? What mass of the Iron will form?

Answer 1

The limiting reactant is magnesium. The theoretical yield of iron is `"31.13 g"`.

Explanation:

Balanced equation

`"Mg(s) + FeCl"_2("aq")` `rarr` `"MgCl"_2("aq") + "Fe(s)"`

The limiting reactant is the reactant that produces the least amount of the product.

In order to determine the limiting reactant, you need to use stoichiometry to determine the mass of iron that each reactant can produce. There are three steps.

1. Determine mol of each reactant by dividing its given mass by its molar mass. Do this by multiplying by the reciprocal of its molar mass.

2. To determine mol iron, multiply mol reactant by the mol ratio between the reactant and iron from the balanced equation, with iron in the numerator.

3. To determine mass of iron, multiply mol iron by its molar mass.

Reactant: Mg

`13.55color(red)cancel(color(black)("g Mg"))xx(1color(red)cancel(color(black)("mol Mg")))/(24.305color(red)cancel(color(black)("g Mg")))xx(1"mol Fe")/(1color(red)cancel(color(black)("mol Mg")))xx(55.845"g Fe")/(1color(red)cancel(color(black)("mol Fe")))="31.13 g Fe"`

Reactant: `"FeCl"_2"`

`80.50color(red)cancel(color(black)("g FeCl"_2))xx(1color(red)cancel(color(black)("mol FeCl"_2)))/(126.745("g FeCl"_2))xx(1color(red)cancel(color(black)("mol Fe")))/(1color(red)cancel(color(black)("mol FeCl"_2)))xx(55.845"g Fe")/(1color(red)cancel(color(black)("mol Fe")))="35.47"`

The limiting reactant in this question is magnesium. The theoretical yield of iron is `"31.13 g"`.