Question

How many grams are in 6.04 x 10^22 particles of AICI3

Answer 1

Answer:  333.415 grams.                                                                                                        This is a very simple chemistry question, which sounds alot like a homework question I did a while ago ;)

Assuming this is a homework question, it doesn't seem that you need to deal with things like impurities, the reactants aren't any special isotopes, you wont be using significant digits, that aluminum is the limiting reactant ect.

anyways, here is how its done.

Your general equation is Al + 3Cl -> AlCl3

A mole of Aluminum weighs 26.98 grams.

26.98 x 2.5 = 67.45 (so that is the weight of the aluminum in your AlCl3 compound.

Now lets add the weight of the Cl.

the molar mass of chlorine is 35.45 grams/mole.

This is where it can get tricky. According to your general equation, you will be using 3 moles of Cl for every mole of Al. So you will actually be calculating the weight of 7.5 moles of Cl.

35.45 x 7.5 = 265.875.

Lets add up the total weight of your product now.

256.875 + 67.45 = 333.415 grams.

Therefore you will have 333.415 grams of AlCl3 if you react 2.5 moles of aluminum completely with Cl2 (a diatomic gas) to create AlCl3

Explanation: