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

Limiting reactant = B2O3

Amount of BCl3 formed = 468 g

Explanation:

The given reaction is:

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

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

false

because it does not have the potential to be entirely used up

yes

explanation:

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