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)

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