Calculating F2 Needed to Produce 120 g of PF3

AI Thread Summary
To calculate the grams of F2 needed to produce 120 g of PF3 with an 84.6% yield, the unbalanced reaction P4(s) + F2(g) → PF3(g) must first be balanced. The user initially divided 120 g by the molar mass of PF3 and adjusted for yield, resulting in 1.612 moles of F2, but this was incorrect due to the oversight of the unbalanced equation. The correct approach requires balancing the equation before calculating the required moles of F2. This highlights the importance of ensuring chemical equations are balanced in stoichiometric calculations. Proper balancing is essential for accurate results in yield-related problems.
Pengwuino
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Ok so this problem pissed me off to no end... basically because ih ad 5 minutes to do it and couldn't figure it out.

Consider the following unbalanced reaction:

P4(s) + F2(g) -----> PF3(g)

How many grams of F2 are needed to produce 120. g of PF3 if the reaction has a 84.6 % yield?

Now, i divided the 120g by the molar mass of PF3 and divided by .846 to find the actual amount necessary which gave me 1.612449659 moles. Multply that by the molar mass of F2 and i get 61.26663726. Wasn't correct, wasn't even a matter of significant figures... what went wrong here :D
 
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Did you remember to balance the equation?
 
oh damn it. They were constantly giving me pre-balanced equations and having to do this so i never noticed that they gave me an unbalanced equation...
 
It says it right there jackass :wink:
 

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