Question

B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)

Calculate how many moles of BCl3 are produced from the reaction of 6122.0g of B203?


A) 3.5 mol

B) 0.0150 mol

C) 211 mol

D) 51.3 mol

Answer 1

Answer is: 175.86 moles of boron(III chloride are produced.
There is some mistake in question.

Balanced chemical reaction:
B₂O₃(s) + 3C(s) + 3Cl₂(g) → 2BCl₃(g) + 3CO(g).
m(B₂O₃) = 6122.0 g.
n(B₂O₃) = m(B₂O₃) ÷ M(B₂O₃).
n(B₂O₃) = 6122 g ÷ 69.6 g.
n(B₂O₃) = 87.93 mol.
From chemical reaction: n(B₂O₃) : n(BCl₃) = 1 : 2.
n(BCl₃) = 2 · 87.93 mol.
n(BCl₃) = 175.86 mol.

Answer 2

Molar mass of B2O3 IS 69.62 g/mol
Therefore; the moles of Boric oxide will be given by 6122/69.62 = 87.94 moles
Moles of B2O3 = 87.94 moles
The mole ratio of B2O3 to BCl3 IS 1:2;
Thus, the moles of BCl3 will be; 87.94 ×2 = 175.9 moles
Hence the moles of BCl3 produced will be; 175.9 moles