Question

Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.16 g/mol.

BCl₃(g) + 3 H2O(l) ? H₃BO₃(s) + 3HCl(g)

I know how to do the math. So Please explain: How do I know which is the limiting reactan

Answer 1

BCl3(g) + 3 H2O(l) ? H3BO3(s) + 3HCl(g)
no.of moles of BCl3 = 60/117.17 = 0.512 moles
no.of moles of H2O = 37.5/18 = 2.08 moles
for 1 mole of BCl3 ......3 moles of H2O requires
so BCl3 is limiting reactent
1mole of BCl3 .......... 3 moles of HCl
0.512 moles of BCl3 can give ..........?
     = 1.536 moles of HCl

mass of HCl formed = 1.536*36.5 = 56.064 gm.