150.0 grams of asf3 were reacted with 180.0 g of ccl4 to produce ascl3 and ccl2f2. the theoretical yield of ccl2f2 produced, in moles, is

The chemical reaction would be written as 

2 AsF3 + 3 CCl4 = 2 AsCl3 + 3 CCl2F2

We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.

150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4

 Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:

1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2