Question

Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants) and iron. 3Mg(s) + 2FeCl3(s) → 3MgCl2(s) + 2Fe(s) A mixture of 41.0 g of magnesium ( = 24.31 g/mol) and 175 g of iron(III) chloride ( = 162.2 g/mol) is allowed to react. What mass of magnesium chloride = 95.21 g/mol) is formed?

Answer 1

Answer:

154.0831 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $Mg$

Given mass = 41.0 g

Molar mass of $Mg$ = 24.31 g/mol

Moles of $Mg$ = 41.0 g / 24.31 g/mol = 1.6865 moles

Given: For $FeCl_3$

Given mass = 175 g

Molar mass of $FeCl_3$ = 162.2 g/mol

Moles of $FeCl_3$ = 175 g / 162.2 g/mol = 1.0789 moles

According to the given reaction:

$3Mg_{(s)}+2FeCl_3_{(s)}\rightarrow 3MgCl_2_{(s)}+2Fe_{(s)}$

3 moles of Mg react with 2 moles of $FeCl_3$

1 mole of Mg react with 2/3 moles of $FeCl_3$

1.6865 mole of Mg react with (2/3)*1.6865 moles of $FeCl_3$

Moles of $FeCl_3$ = 1.1243 moles

Available moles of $FeCl_3$ = 1.0789 moles

Limiting reagent is the one which is present in small amount. Thus, $FeCl_3$ is limiting reagent. (1.0789 < 1.1243)

The formation of the product is governed by the limiting reagent. So,

2 moles of $FeCl_3$ gives 3 moles of magnesium chloride

1 mole of $FeCl_3$ gives 3/2 moles of magnesium chloride

1.0789 mole of $FeCl_3$ gives (3/2)*1.0789 moles of magnesium chloride

Moles of magnesium chloride = 1.61835 moles

Molar mass of magnesium chloride = 95.21 g/mol

Mass of magnesium chloride = Moles × Molar mass = 1.61835 × 95.21 g = 154.0831 g