Question

Magnesium reacts with iron chloride solution.

3 Mg + 2 FeCl3 ⟶ 2 Fe + 3 MgCl2

0.120 g of magnesium reacts with excess iron chloride solution.

Relative atomic masses (Ar): Mg = 24 Fe = 56


Calculate the mass of iron produced, in mg
Show your working.

Answer 1

Answer:

0.1876g

Explanation:

Since iron is in excess, therefore Magnesium is the limiting reactant.

Moles of Mg = mass/molar mass

0.120/24 = 0.005 moles

Compare the coefficient ratio of Mg and iron.

Ratio is 3 : 2

To find moles of iron we take 2/3 x 0.005

Moles of iron is 0.00335 moles

Mass = moles x molar mass

= 0.00335 x 56

= 0.1876g

Answer 2

Taking into account the reaction stoichiometry, 0.186 grams of Fe are produced when 0.120 g of magnesium reacts with excess iron chloride solution.

Reaction stoichiometry

In first place, the balanced reaction is:

3 Mg+ 2 FeCl₃  → 2 Fe+ 3 MgCl₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• Mg: 3 moles
• FeCl₃: 2 moles  
• Fe: 2 moles
• MgCl₂: 3 moles

The molar mass of the compounds is:

• Mg: 24 g/mole
• FeCl₃: 162.35 g/mole
• Fe: 56 g/mole
• MgCl₂: 94.9 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• Mg: 3 moles ×24 g/mole= 72 grams
• FeCl₃: 2 moles ×162.35 g/mole= 324.7 grams
• Fe: 2 moles ×56 g/mole= 112 grams
• MgCl₂: 3 moles ×94.9 g/mole= 284.7 grams

Mass of iron produced

The following rule of three can be applied: if by reaction stoichiometry 72 grams of Mg form 112 grams of Fe, 0.120 grams of Mg form how much mass of Fe?

$mass of Fe=\frac{0.120 grams of Mgx112 grams of Fe}{72 grams of Mg}$

mass of Fe= 0.186 grams

Then, 0.186 grams of Fe are produced when 0.120 g of magnesium reacts with excess iron chloride solution.

Learn more about the reaction stoichiometry:

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