Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. Select one: a. Limiting reactant is Mg; 7.4 g of FeCl3 remain. b. Limiting reactant is Mg; 46.5 g of FeCl3 remain. c. Limiting reactant is FeCl3; 1.7 g of Mg remain. d. Limiting reactant is FeCl3; 37.8 g of Mg remain. e. Limiting reactant is Mg; 134.0 g of FeCl3 remain.

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2021-05-11T21:45:17+02:00

Answer:

Limiting reactant is FeCl3; 1.7 g of Mg remain.

Explanation:

From the question;

The equation is;

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

Amount of Mg = 41 g/24.31 g/mol = 1.687 moles

The limiting reactant yields the least amount of MgCl2

3 moles of Mg yields 3 moles of MgCl2

Hence 1.687 moles of Mg yields yields 1.687 moles of MgCl2

FeCl₃ = 175 g/162.2 g/mol = 1.0789 moles

2 moles of FeCl3 yields 3 moles of MgCl2

1.0789 moles of FeCl3 yields 1.0789 * 3/2 = 1.61835 moles of MgCl2

Hence FeCl3 is the limiting reactant

3 moles of Mg reacts with 2 moles of FeCl3

x moles of Mg reacts with 1.0789 of FeCl3

x = 3 * 1.0789 /2 = 1.61835 moles of Mg

Mass of Mg reacted = 1.61835 moles * 24.31 = 39.342 g

Mass of excess Mg = 41 g - 39.342 g = 1.7 g