1.- Magnesium(used in the manufacture of light alloys) reactswith iron(III) chlo

Question

1.- Magnesium(used in the manufacture of light alloys) reactswith iron(III) chloride to form magnesium chloride and iron. Amixture of 41.0 g of magnesium and 175 g of iron(III) chloride isallowed to react. Identify the limiting reactant and determine themass of the excess reactant present in the vessel when the reactionis complete.     a) Limiting reactant is Mg: 67g of FeCl3 remains.                                     b)Limiting reactant is Mg: 134 g FeCl3remains.                                    c) Limiting reactant is Mg: 104 g FeCl3remains.                                    d) Limiting reactant is FeCl3: 1.7 g of Mg remans.                                     e)Limiting reactant is FeCl3: 87.2 g of Mg remains. 2.- What is the % of oxygen in iron(III) sulfate?  a)28% b)32% c)24% d)48% e)42%                                     1.- Magnesium(used in the manufacture of light alloys) reactswith iron(III) chloride to form magnesium chloride and iron. Amixture of 41.0 g of magnesium and 175 g of iron(III) chloride isallowed to react. Identify the limiting reactant and determine themass of the excess reactant present in the vessel when the reactionis complete.     a) Limiting reactant is Mg: 67g of FeCl3 remains.                                     b)Limiting reactant is Mg: 134 g FeCl3remains.                                    c) Limiting reactant is Mg: 104 g FeCl3remains.                                    d) Limiting reactant is FeCl3: 1.7 g of Mg remans.                                     e)Limiting reactant is FeCl3: 87.2 g of Mg remains. 2.- What is the % of oxygen in iron(III) sulfate?  a)28% b)32% c)24% d)48% e)42%                                    

Explanation / Answer

3Mg + 2FeCl3   =>   3MgCl2   +2Fe 41.0 g Mg * (1mol/24.3 grams) *( 2 mol FeCl3/3 mol Mg) = 1.125moles FeCl3 required actual amount: 175 g FeCl3 * (1mol/162.2 grams ) = 1.07 moles of FeCl3 actuallyavailable. Thus, FeCl3 is the limiting reactant 175 g FeCl3*(1mol FeCl3/162.2 grams) *(3 mol Mg/2 mol FeCl3) *(24.3g Mg /1mol Mg) = 39.33 g Mg reacted 41.0 g - 39.33 g = 1.67 grams ~ 1.7 grams left Answer is d)

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