1.- Magnesium(used in the manufacture of light alloys) reactswith iron(III) chlo
ID: 76012 • Letter: 1
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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