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
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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