Question

What is my actual yield of uranium hexabromide if start with 100 grams of uranium and get a percent yield of 83%?

`U + 3Br_2 -> UBr_6`

Answer 1

Assuming that uranium is the limiting reactant, and the equation you stated is correct,

`U+3Br_2 to UBr_6`

The theoretical yield is,

`100g * (U)/(238g) * (UBr_6)/(U) approx 0.420"mol"`,

and if the percent yield is `83%`,

`0.83 = (A)/(0.420)`
`therefore A approx 0.350"mol"`

is the actual yield of uranium hexabromide given this data.