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

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 

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.

The answer is b metal denting