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

Answer:

Limiting reactant = B2O3

Amount of BCl3 formed = 468 g

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

$Moles(B2O3)=\frac{Mass(B2O3)}{Mol.wt(B2O3)}=\frac{139g}{69.6g/mol}=1.997moles$

$Moles(C)=\frac{Mass(C)}{At.wt(C)}=\frac{87.8g}{12g/mol}=7.317moles$

$Moles(Cl2)=\frac{Mass(Cl2)}{Mol.wt(Cl2)}=\frac{650g}{70.9g/mol}=9.168moles$

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

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