For this question we have to start with balanced chemical equation.
U + 3
As per the above equation (a) , one mole of Uranium U, produces one mole of U
1 mole of reactant Uranium = 1 mole of U
In terms of mass one mole of Uranium has mass 238.02 g/mol.
and one mole of U
so let us set up the ratio;
(1 mole U / 1 mole U
100 g of U will produce X g of U
100 g of U / X g of U
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
0.83 x 301.40 = 250.16 g.