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?

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

Answer 1

I bet this is a high temperature process. I get a possible yield of `88*mol` with respect to `"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?