Question

Consider the reaction of diboron trioxide with carbon and chlorine. B2O3 (s) + 3C (s) + 3Cl2 (g) 2BCl3 (g) + 3CO (g) Determine the limiting reactant in a mixture containing 139 g of B2O3, 87.8 g of C, and 650 g of Cl2. Calculate the maximum mass (in grams) of boron trichloride, BCl3, that can be produced in the reaction. The limiting reactant is: B2O3 Cl2 C Amount of BCl3 formed = g

Answer 1

Limiting reactant = B2O3

Amount of BCl3 formed = 468 g

Step-by-step explanation:

The given reaction is:

`B2O3 (s) + 3C (s) + 3Cl2 (g) \rightarrow 2BCl3 (g) + 3CO (g)`

In order to identify the limiting reagent calculate the moles of B2O3, C and Cl2. The reagent with the lowest moles is the limiting reactant

`Moles(B2O3)=(Mass(B2O3))/(Mol.wt(B2O3))=(139g)/(69.6g/mol)=1.997moles`

`Moles(C)=(Mass(C))/(At.wt(C))=(87.8g)/(12g/mol)=7.317moles`

`Moles(Cl2)=(Mass(Cl2))/(Mol.wt(Cl2))=(650g)/(70.9g/mol)=9.168moles`

Since the moles of B2O3 < C < Cl2, the limiting reactant is B2O3

Based on the reaction stoichiometry:

1 mole of B2O3 produces 2 moles of BCl3

Hence, the number of moles of BCl3 produced under the experimental conditions = 2*1.997=3.994 moles

`Mass(BCl3)= Moles* Mol.wt = 3.994 moles*117.17g/mol = 468 g`