posted by lyne on .
Consider the following unbalanced reaction.
P4(s) + 6F2(g) 4PF3(g)
How many grams of F2 are needed to produce 114. g of PF3 if the reaction has a 78.1% yield?
i tried it and got this
114g PF3 x (mol PF3/87.97g PF3) x (6 mol F2/ mol PF3) x (38g F2/mol F2)=73.9gF2
then plugged it into the % yield equation and got 57.7
what did i do wrong??
You're ok as far as you went but you didn't do the % yield right.
If the yield is only 78.1%, then you know that you must take MORE, not less than 73.9.
73.9/0.781 = ??