How many grams of iron are needed to react with 31.0 L of chlorine gas at STP to produce iron(III) chloride?Please show your work and use Stoichiometry to solve the problem.

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pacorz | High School Teacher | (Level 3) Educator

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You have 31 Liters if Cl2 given. The ideal gas law says that 22.4 Liters of any gas at STP is equal to one mole, so our conversion looks like this for step 1:

31L Cl2/1  x  1 Mole Cl2/ 22.4 L Cl2  = 1.384 Moles Cl2

Iron III Chloride's formula is Fe Cl3, since the iron gives away 3 electrons per atom and each Chlorine atom can only accept 1 electron.

We know that each Cl2 molecule provides 2 Cl atoms, and that for each Fe atom we will need 3 Chlorine atoms. Combining those facts gives us this conversion for step 2:

1.384 M Cl2/1 x 2 M Cl/1 M Cl2 x 1 M Fe/3 M Cl  = .923 Moles Fe

1 Mole of Fe weighs 55.84 grams, so step three gets us to grams this way:

,923 Moles Fe/1 x 55.84 g Fe/1 Mole Fe = 51.52 grams Fe