Answer
The max yield for this reaction is equal to 401g.
Work Step by Step
Determine the limiting reactant.
1. Calculate the number of moles of $UO_3$:
238.03* 1 + 16* 3 = 286.03g/mol
$365g \times \frac{1 mol}{ 286.03g} = 1.276mol (UO_3)$
- According to the balanced equation:
The ratio of $UO_3$ to $BrF_3$ is 6 to 8:
$1.276 mol (UO_3) \times \frac{ 8 mol(BrF_3)}{ 6 mol (UO_3)} = 1.701mol (BrF_3)$
2. Calculate the mass of $BrF_3$:
79.9* 1 + 19.00* 3 = 136.9g/mol
$1.701 mol \times \frac{ 136.9 g}{ 1 mol} = 232.9g (BrF_3)$
Therefore, we have more $BrF_3$ (365g) than we need (232.9g), making it the excess reactant and $UO_3$ the limiting one.
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Use the limiting reactant to find the mass of the products.
According to the balanced equation:
The ratio of $UO_3$ to $UF_4$ is 6 to 6.
$1.276 mol (UO_3) \times \frac{ 6 mol(UF_4)}{ 6 mol (UO_3)} = 1.276mol (UF_4)$
3. Calculate the mass of $UF_4$:
238.03* 1 + 19.00* 4 = 314.03g/mol
$1.276 mol \times \frac{ 314.03 g}{ 1 mol} = 401g (UF_4)$
