150.0 g of AsF3 was reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in moles, is? a]0.7802 mol b]1.274 mol c]1.170 mol d]0.5685 mol e]1.705 mol

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150.0 g of AsF3 was reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in moles, is? a]0.7802 mol b]1.274 mol c]1.170 mol d]0.5685 mol e]1.705 mol

Mathematics
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AsF3 + CCl4 = AsCl3 + CCl2F2 you need all the mol number of each substance of both sides to be equal. start with F 3F on the left, 2F on the right so time 2 to 3F on the left, time 3 to 2F com the right, we get 2AsF3 + CCl4 = AsCl3 + 3CCl2F2 Now, there are 2 As on the left, 1 As on the right so time 2 to As on the right to balance 2AsF3 + CCl4 = 2AsCl3 + 3CCl2F2 now there are 4Cl left, 12Cl right, also 1C left, 3C right so time 3 to CCl4 2AsF3 + 3CCl4 = 2AsCl3 + 3CCl2F2 that's the balance part
when u get used to it, u will just put number in there by trying, shouldn't take more than 30 secs.
Wow... what do I do afterwards?

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Now group what information you have M: atomic mass, u know this. n: mol number 150.0 g of AsF3 MAs=75g/mol MF=19g/mol MAsF3=75+19*3=132g/mol nAsF3=150/132=1.136mol 180.0 g of CCl4 MC=12 MCl=35.5 MCCl4=12+35.5*4=154 nCCl4=180/154=1.169mol
Thank you so much :) Now you can go back to bed :P
2AsF3 + 3CCl4 = 2AsCl3 + 3CCl2F2 theoretically 2mol 3mol don't need 3mol reality 1.136 1.169 should be 1.132 -> 1.704 try with CCl4 1.169 blabla, so you can take over from here right? i'll leave the rest for you
no prob

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