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

Answer 2

The mass of magnesium chloride (MgCl₂) formed is 154.08 g

From the question,

The balanced chemical equation for the reaction is

3Mg(s) + 2FeCl₃(s) → 3MgCl₂(s) + 2Fe(s)

This means 3 moles of Mg reacts with 2 moles of FeCl₃ to produce 3 moles of MgCl₂ and 2 moles of Fe

To determine the mass of magnesium chloride formed,

First, we will determine the number of moles of the reactants that are present.

• For Magnesium (Mg)

Mass = 41.0 g

Molar mass = 24.31 g/mol

From the formula,

$Number \ of \ moles = \frac{Mass}{Molar\ mass}$

∴ $Number\ of\ moles\ of\ Magnesium = \frac{41.0}{24.31}$

Number of moles of Magnesium = 1.6865 moles

• For Iron(III) chloride (FeCl₃)

Mass = 175 g

Molar mass = 162.2 g/mol

∴ $Number\ of\ moles\ of\ Iron(III) chloride = \frac{175}{162.2}$

Number of moles of Iron(III) chloride = 1.0789 moles

Since  3 moles of Mg reacts with 2 moles of FeCl₃

Then,

$\frac{3}{2}\times 1.0789$ moles of Mg will react with 1.0789 moles of FeCl₃

$\frac{3}{2}\times 1.0789 = 1.61835$ moles

This means only 1.61835 moles of Mg reacted

(NOTE: Magnesium is the excess reagent while Iron(III) chloride is the limiting reagent)

From the balanced chemical equation for the reaction

3 moles of Mg reacts with 2 moles of FeCl₃ to produce 3 moles of MgCl₂

Then,

1.61835 moles of Mg will react with 1.0789 moles of FeCl₃ to produce 1.61835 moles of MgCl₂

∴ 1.61835 moles of MgCl₂ is produced during the reaction

Now, for the mass of magnesium chloride formed,

Number of moles of magnesium chloride = 1.61835 moles

Molar mass of magnesium chloride = 95.21 g/mol

Using the formula

Mass = Number of moles × Molar mass

Mass of magnesium chloride formed =  1.61835 × 95.21

Mass of magnesium chloride formed = 154.08 g

Hence, the mass of magnesium chloride (MgCl₂) formed is 154.08 g

Learn more here: https://brainly.com/question/13691730