## Chemistry: The Molecular Science (5th Edition)

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. ------------------------------------- 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 $1.276 mol \times \frac{ 314.03 g}{ 1 mol} = 401g (UF_4)$