Ok, this is a tricky one with some really crazy calculations if you can't use a calculator, where did you get this problem from? You probably won't see calculations like this on the actual test, but here's how I did it.

Step #1 - So you've got 7.18 grams of AgCl, add the atomic masses togeather and you get 143g per mole of AgCl. Now divide 7.18g/143g = 0.05 moles of AgCl.

Step #2 - Knowing that you have 0.05 moles of AgCl, and that there are 3 moles of Cl in VOCl3 and all of it went to make AgCl, divide 0.05 moles by 3 to get 0.0167 moles of VOCl3.

Step #3 - Since we know there are 2.8934g of pure VOCl3, set up the equation like this:

2.8934g VOCl3 = 0.0167 moles VOCl3, then divide both sides by 0.0167 to get that 1 mole of VOCl3 is 173.26g.

Step #4 - Since we know the molar mass of Cl (35.453) and O (16), just add up what you have, and the molar mass of V is the difference from 173.26.

So, V + 16 + (35.453*3) = 173.26 , solve for V and you get that V = 50.9

Not too bad, but this would take you way too long without a calculator. If you got a question like this on the test, I'd start rounding like crazy.