# Question #b7af7

Mar 15, 2014

U + 3$B {r}_{3}$ --------> U$B {r}_{6}$ (a)

As per the above equation (a) , one mole of Uranium U, produces one mole of U$B {r}_{6}$

1 mole of reactant Uranium = 1 mole of U$B {r}_{6}$.

In terms of mass one mole of Uranium has mass 238.02 g/mol.
and one mole of U$B {r}_{6}$ has mass 717.4 g/mol.

so let us set up the ratio;

(1 mole U / 1 mole U$B {r}_{6}$ ) = 238.02 g of U / 717.4 g of U$B {r}_{6}$ (b)

100 g of U will produce X g of U$B {r}_{6}$.

100 g of U / X g of U$B {r}_{6}$ (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 percent yield is 83 % , so the reaction produced 83 % of the expected yield. The actual amount of U$B {r}_{6}$ is

0.83 x 301.40 = 250.16 g.