How many grams of FeCl3 are produced when 426 grams of chlorine react with iron, according to this equation?
2Fe + 3Cl2 → 2FeCl3
How many moles of FeCl3 will be produced when we start with 6 moles of Cl2?

Question

How many grams of FeCl3 are produced when 426 grams of chlorine react with iron, according to this equation? 
2Fe  +  3Cl2   →  2FeCl3
How many moles of FeCl3 will be produced when we start with 6 moles of Cl2?

micahm · Answer

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anonymous · Answer

I'm assuming (but correct me if I'm wrong) that Fe is in excess.

Can you convert 426 grams of chlorine to moles?

jebonna · Answer

To answer these questions, its always easy to remember the triangle of equations for working out mass, moles and molar masses of substances: 
|dw:1449184564943:dw|
Where m = mass (g)
M = molar mass (g mol-1)
n = moles (mol)

To work out what to do to find out what you need, just cover up the desired product with your finger, and then do whatever the triangle tells you to. 

For example, if I wanted to find out how many moles a substance has, I would cover the n (which stands for moles) and do m/M (mass divided by molar mass), and that will give me the amount of moles.

So, to work out your first question, you will have to work out the amount of moles of chlorine (Use triangle). Then, you will have to use the stoichiometry in the equation to work out the number of moles of FeCl3 (There is a '3' in front of chlorine, and a '2' in front of FeCl3. The answer you will have will be for TWO lots of Cl2, and you need the answer for THREE lots of FeCl3. You have to change the answer from 2 lots to 3 lots) And lastly, you will have to find the mass of FeCl3 (Use triangle).

To work out your second question, its all to do with stoichiometry again. 6mol is for THREE lots of Cl2. You need TWO lots of FeCl3. The easiest way to do this is to divide your answer (6mol) by 3 (to get ONE lot of a substance), and then times your answer by 2 (to get TWO lots of a substance). That will be your answer. 

I hope this makes sense, and helps! :)