Magnesium (used in the manufacture of light alloys) reacts with iron(III) chlood

Question

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloode to form magnesium chtonde and iron. A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride Is allowed to react according to the following balanced equation. 3Mg + 2FeCl_3 rightarrow 3MgCl_2 + 2Fe Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. Limiting reactant is FeCl_3; 37.8 g of Mg remain. Limiting reactant is FeCl_3; 1.7 g of Mg remain. Limiting reactant is Mg; 134.0 g of FeCl_3 remain. Limiting reactant is Mg; 7.4 g of FeCl_3 remain. Limiting reactant is Mg; 46.5 g of Fecl_3 remain.

Explanation / Answer

Mass of Magnesium = 41 gm mass of FeCl3= 175 gm

The reaction is 3Mg+ 2FeCl3------------->3MgCl2+ 2Fe

Atomic weights : Mg= 24.3, Fe= 56 and chlorine= 35.5

Molecular weights : FeCl3= 56+3*35.5=162.5, MgCl2= 24+71= 95 and Fe=56

given Moles of Mg= 41/24.3=1.687 and moles of FeCl3= 175/162.5=1.08

Molar ratio of Mg :FeCl2= 1.687: 1.08 = 1.56 : 1

As per the reation given, molar ratio of Mg : FeCl2 = 3: 2 = 1.5 :1

So Mg is excess reagent. FeCl3 is limiting reactant

2 moles of FeCl3 reacts wtih 3 moles of Mg

1.08 moles of FeCl3 requires 1.08*1.5=1.62 moles of Mg

excess mole of Mg= 1.687-1.62= 0.06=0.067 moles

Mass of Mg remaining = 1.63 gms   Limiting reactant is FeCl3 and close answer is 1.7 Mg

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