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

Calculate how many moles of BCl3 are produced from the reaction of 61.0g of B203?
A. 1.75 mol
B. 0.0150 mol
C. 51.3 mol
D. 211 mol