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QUESTION

# 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 + 3Br_3 --------> UBr_6 (a)

As per the above equation (a) , one mole of Uranium U, produces one mole of UBr_6

1 mole of reactant Uranium = 1 mole of UBr_6.

In terms of mass one mole of Uranium has mass 238.02 g/mol. and one mole of UBr_6 has mass 717.4 g/mol.

so let us set up the ratio;

(1 mole U / 1 mole UBr_6 ) = 238.02 g of U / 717.4 g of UBr_6 (b)

100 g of U will produce X g of UBr_6.

100 g of U / X g of UBr_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 UBr_6 is

0.83 x 301.40 = 250.16 g.