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 = Â 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
The reaction yields 64.15 grams HCl.
According to the given equation,
[tex]\rm BCl_3\;+\;3\;H_2O\;\rightarrow\;H_3BO_3\;+\;3\;HCl[/tex]
1 mole of [tex]\rm BCl_3[/tex] yields 3 moles of HCl.
117.16 grams of [tex]\rm BCl_3[/tex] yields 3 [tex]\times[/tex] 36.5 grams of HCl.
So,
117.16 grams [tex]\rm BCl_3[/tex] = 109.5 grams HCl
60.0 grams [tex]\rm BCl_3[/tex] = [tex]\rm \frac{117.16}{109.5}\;\times\;60[/tex]
= 64.15 grams of HCl.
The reaction yields, 64.15 grams of HCl.
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