B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)

Question

How many moles of chlorine are used up in a reaction that produced 0.35 kg of BCl3

I do not understand please help

Answer 1

With some minor variations, all of these stoichiometry problems are alike.
Step 1. Write and balance the equation. You have that.

Step 2. Convert what you have to mols. That is mols = grams/molar mass = 350g/117 = approx 3 but you need to confirm all of these estimates here and those that follow.

Step 3. Using the coefficients in the balanced equation, convert mols of what you have (in this mols BCl3) to mols of what you want (in this case mols Cl2). That is done this way.
3 mols BCl3 x 3 mols Cl2/2 mol BCl3) = 3 x 3/2 = 9/2 = 4.5 mols Cl2. The problem stops here; i.e., the problem asks for mols Cl2.

Step 4. The usual question asks for grams Cl2. To find grams Cl2 it's g Cl2 = mols Cl2 x molar mass Cl2 = 4.5 x (2*35.44) = ?

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