If 98.0 g B₂O₃ (69.62 g/mol)and 125 g HF (20.01 g/mol)are combined and allowed to react according to the given equation,what mass of the excess reactant will be left over? B₂O₃(s)+ 6 HF(aq)$\to$ 2 BF₃(g)+ 3 H₂O

Question

Quizplus · Accepted Answer

1 mol B₂O₃ has a mass of 69.62 g. 98.0 g B₂O₃ is equal to 1.41 mol. 1 mol HF has a mass of 20.01 g. 125 g HF is equal to 6.25 mol. According to the balanced chemical equation, 1 mol B₂O₃ reacts with 6 mol HF. The limiting reactant in this case is B₂O₃, as we have only 1.41 mol of it, which is not enough to react with all 6.25 mol HF. Using stoichiometry, we can calculate the amount of HF required to react with all 1.41 mol of B₂O₃: 1 mol B₂O₃ x 6 mol HF/1 mol B₂O₃= 6 mol HF Therefore, the amount of excess HF left over is: 6.25 mol HF - 6 mol HF = 0.25 mol HF Converting this to grams: 0.25 mol HF x 20.01 g/mol = 5.0 g HF To find the mass of excess B₂O₃ left over, we can use stoichiometry again: 1 mol B₂O₃ x 6 mol HF/1 mol B₂O₃= 6 mol HF This means that to react with all 6.25 mol HF, we would need: 6.25 mol HF x 1 mol B₂O₃/6 mol HF = 1.04 mol B₂O₃ Converting this to grams: 1.04 mol B₂O₃ x 69.62 g/mol = 72.3 g B₂O₃ Subtracting the mass of B₂O₃ used from the total mass we started with: 98.0 g - 72.3 g = 25.7 g Therefore, the mass of excess B₂O₃ left over is 25.7 g. The correct answer is B, 25.5 g B₂O₃ will be left over.

If 980 G B2O3 (69 →\to→ 2 BF3(g)+ 3 H2O A)72

If 980 G B₂O₃ (69 $\to$ 2 BF₃(g)+ 3 H₂O
A)72