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) + 3 HCl(g)

Answer 1

   The   theoretical yield  of  HCl  produced  is  56.06  grams  of HCl

    calculation

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

Step 1: find the  moles of   each reactant

 moles  = mass÷molar mass

moles  of BCl₃  =  60.0 g÷117.16 g/mol =0.512  moles

moles  of H2O  = 37.5 g ÷18 g/ mol = 2.083  moles

Step 2: use the  moles  ratio  to determine  the limiting reagent

from  the  equation above BCl₃ :HCl  is 1:3 therefore the moles of HCl =  0.512 moles x 3/1 =1.536  moles

H2O :HCl  is 3:3 = 1:1 therefore  the moles of  HCl is also 2.083  moles

Bcl₃  is  the limiting   reagent since it produces  less  amount  of HCl   therefore the   moles   HCl is 1.536  moles

Step 3 ; find the  theoretical  yield

The theoretical yield =  moles x molar mass

from periodic table the molar mass  of HCl = 1+ 35.5=  36.5 g/mol

Theoretical yield= 1.536 moles  x  36.5 g/mol =56.06 g  of Hcl