Note your typo at the beginning. You obtain 10.9 g SbI3, This is a limiting reagent (LR) problem. I do these the long way.
2Sb + 3I2 -> 2SbI3
mols Sb = grams/atomic mass Sb = 8.60/121.8 = estimated 0.07
mols I = 11,4/126.9 = estimated 0.09mols SI3 formed
mols SbI3 formed with Sb and excess I2
0.07 x (2 mols SbI3/2 mols Sb) = 0.07
mols SbI3 formed from 0.09 mols I2 and excess Sb,
0.09 mols I2 x (2 mols SbI3/3 mols I2) = 0.06
Since I2 will form fewer mols SbI3 then I2 is the LR and Sb is the excess reagent (ER)
grams SbI3 formed = mols SbI3 x molar mass SbI3 = ? This is the theoretical yield (TY) which assumes it is 100% efficient.
The actual yield (AY) from the problem is 10.9 grams.
% yield = (AY/TY)*100 = ?
Note that I've estimated the math calculations so you should repeat all of them. Post your work if you get stuck. You might want to confirm the atomic masses to make sure I looked them up properly.
if you react 8.60g of Sb with 11.4g of I and recover 10.9 of SbI, what is the percentage yield of antimony iodide
2Sb + 3I2 -> 2SbI3
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