What is your actual yield of uranium hexabromide if you start with 100 grams of uranium and get a percent yield of 83% ? U + 3 Br2 → UBr6

For this question we have to start with balanced chemical equation.

U + 3##Br_3## --------> U##Br_6## (a)

As per the above equation (a) , one mole of Uranium U, produces one mole of U##Br_6##

1 mole of reactant Uranium = 1 mole of U##Br_6##.

In terms of mass one mole of Uranium has mass 238.02 g/mol. and one mole of U##Br_6## has mass 717.4 g/mol.

so let us set up the ratio;

(1 mole U / 1 mole U##Br_6## ) = 238.02 g of U / 717.4 g of U##Br_6## (b)

100 g of U will produce X g of U##Br_6##.

100 g of U / X g of U##Br_6## (c)

equating two equation (b) and (c)

238.02 / 717.4 = 100 g / X

238.02 . X = 717.4 x 100

238.02 . X = 71740

X = 71740/ 238.02 = 301.40 g

since the is 83 % , so the reaction produced 83 % of the expected yield. The actual amount of U##Br_6## is

0.83 x 301.40 = 250.16 g.