Magnesium metal react beside iron(III) chloride to form magnesium chloride and iron.?
A mixture of 41.0g of Mg (molar mass= 24.31g/mol) and 175g of iron(III) chloride (molar mass = 162.2 g/mol) is allowed to react/ Identify the limiting reactant and determine the mass of the excess reactant present present in the vessel when the reaction is complete.
3Mg (s) + 2FeCl3 (s) ---------> 3MgCl2 (s) + 2Fe (s)
* please sustain with this problem, provide work and explanations if possible, gratitude!
Answers:
To answer the question, first express all the bulk in terms of moles. A mole is newly a number chemists use to express weight. This many atoms of any item equals its atomic weight in the broken up table.
41 g of Mg is 41 g / (24.31 g/mole) = 1.67 moles of Mg.
Next, note the ratios of atoms within the chemical equation. It must be balanced, meaning here are equal numbers of atoms on each side.
3Mg + 2FeCl3 --> 3MgCl2 + 2Fe
3 atoms of Mg will require 2 molecules of FeCl3 to react completely. This is also true of the molar ratio too (since both are just numbers). This means the ratio of FeCl3 to Mg is 2 to 3. If 1.67 moles of Mg are present, they will form 1.67 moles Mg / (2 moles Mg / 3 moles Fecl3) = 1.67 / (2/3) = 2.53 moles FeCl3.
175 grams of FeCl3 is 175 grams / (162.2 grams / mole) = 1.08 moles FeCl3. However, it be calculated that 2.53 moles are needed. This means 2.53 - 1.08 = 1.45 moles of FeCl3 remain after the reaction.
because not plenty FeCl3 was present to completely react, it is the limiting reagent. The remaining mass is the number of moles of FeCl3 times the molecular immensity of FeCl3. Source(s): I'm a chemist!
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3Mg (s) + 2FeCl3 (s) ---------> 3MgCl2 (s) + 2Fe (s)
* please sustain with this problem, provide work and explanations if possible, gratitude!
Answers:
To answer the question, first express all the bulk in terms of moles. A mole is newly a number chemists use to express weight. This many atoms of any item equals its atomic weight in the broken up table.
41 g of Mg is 41 g / (24.31 g/mole) = 1.67 moles of Mg.
Next, note the ratios of atoms within the chemical equation. It must be balanced, meaning here are equal numbers of atoms on each side.
3Mg + 2FeCl3 --> 3MgCl2 + 2Fe
3 atoms of Mg will require 2 molecules of FeCl3 to react completely. This is also true of the molar ratio too (since both are just numbers). This means the ratio of FeCl3 to Mg is 2 to 3. If 1.67 moles of Mg are present, they will form 1.67 moles Mg / (2 moles Mg / 3 moles Fecl3) = 1.67 / (2/3) = 2.53 moles FeCl3.
175 grams of FeCl3 is 175 grams / (162.2 grams / mole) = 1.08 moles FeCl3. However, it be calculated that 2.53 moles are needed. This means 2.53 - 1.08 = 1.45 moles of FeCl3 remain after the reaction.
because not plenty FeCl3 was present to completely react, it is the limiting reagent. The remaining mass is the number of moles of FeCl3 times the molecular immensity of FeCl3. Source(s): I'm a chemist!
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