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?

Does anyone know how to solve this and/or how to find the mole to mole ratio?

1 Answer
Apr 29, 2018

I bet this is a high temperature process. I get a possible yield of 88*mol88mol with respect to "boron trichloride..."boron trichloride...

Explanation:

We gots...

B_2O_3(s) + 3C(s) + 3Cl_2(g) stackrel(Delta)rarr2BCl_3(g) + 3CO(g)

"Moles of boric oxide"=(6122.0*g)/(69.62*g*mol^-1)=87.94*mol

And so at most, we can make TWICE this molar quantity of BCl_3, i.e. 175.9*mol...and what mass of BCl_3 does this represent?