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Report (any type)/Brief report
22/09/2021
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.
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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