Question

For the following reaction, 22.0 grams of iron are allowed to react with 45.3 grams of chlorine gas . iron(s) + chlorine(g) iron(III) chloride(s) . What is the maximum mass of iron(III) chloride that can be formed? ____grams . What is the FORMULA for the limiting reagent?. What mass of the excess reagent remains after the reaction is complete?____ grams

Answer 1

Balanced equation for the reaction:

2Fe(s) + 3Cl₂(g)= 2FeCl₃(s)

According to the stoichiometric equation 2 mol Fe reacts with 3 mol Cl2 to form 2 mol FeCl3

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CALCULATION OF NUMBER OF MOLES OF Fe

Molar mass of Fe = 55.8 g/mol

Number of moles of Fe = Mass of Fe / Molar mass of Fe

                                                         = 22.0 g /55.8 g/mol = 0.394 mol

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CALCULATION OF NUMBER OF MOLES OF Cl2

Molar mass of Cl2 =(2 x 35.5) g/mol = 71.0 g/mol

Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2                                                       

                                     =    45.3 g /71 g/mol = 0.638 mol

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DETERMINATION OF THE LIMITING REAGENT

2Fe(s) + 3Cl2(g)= 2FeCl3(s)

According to the stoichiometric equation 2 mol Fe requires 3 mol Cl2

Therefore 0.394 mol Fe will require (3 x 0.394 mol / 2) mol Cl2 = 0.591 mol Cl2

Number of moles of Cl2 available for the reaction = 0.638 mol

This is more than what is required

Hence, 0.394 mol Fe can react completely.

Therfore the limiting reagent is Fe

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CALCULATION OF THE NUMBER OF MOLES FeCl3 AND HENCE THE MASS OF FeCl3THAT CAN BE FORMED

2Fe(s) + 3Cl2(g)= 2FeCl3(s)

2 mol of Fe gives 2 mol of FeCl3

Therefore

0.394 mol of Fe will give 0.394 mol of FeCl3

Molar mass of FeCl₃= (55.8 + 3 x 35.5) g/mol = 162.3 g/mol

1 mol of FeCl3 = 162.3 g /mol

Therefore 0.394 mol of FeCl3 = 0.394 mol x 162.3 g/mol = 63.9 g

The maximum mass of FeCl3 formed = 63.9 g

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CALCULATION OF THE NUMBER OF MOLES OF EXCESS Cl2 AND HENCE THE NUMBER OF GRAMS OF EXCESS Cl2

Since Fe reacts completely, Cl2 is the excess reagent

0.394 mol Fe requires = 0.591 mol Cl2

The number of moles of Cl2 present initially = 0.638 mol

The number of moles of Cl2 that will remain after the reaction = 0.638 mol - 0.591 mol = 0.0470 mol

1 mol of Cl2 = 71 g/mol

Therefore, 0. 0470 mol of Cl2 = 0. 0470 mol x 71 g/mol = 3.34 g

The number of grams of Cl2 (excess reagent) that will remain after the completion of the reaction = 3.34 g

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