The reaction for the preparation of boric acid from tri halide of boron and water is as follows:

BX3 + 3H2O → B(OH)3 + 3HX

Where BX3 is the tri halide of boron (Boron Trichloride, Boron Tribromide, etc.) and HX is a halogen acid (Hydrochloric acid, Hydrobromic acid, etc.)

From the balanced chemical equation, we can see that 1 mole of BX3 reacts with 3 moles of H2O to produce 1 mole of B(OH)3 and 3 moles of HX.

Therefore, the total number of moles of reactants involved in the reaction is 1 (from BX3) + 3 (from H2O) = 4 moles.

Question

The reaction for the preparation of boric acid from tri halide of boron and water is as follows:

BX3 + 3H2O → B(OH)3 + 3HX

Where BX3 is the tri halide of boron (Boron Trichloride, Boron Tribromide, etc.) and HX is a halogen acid (Hydrochloric acid, Hydrobromic acid, etc.)

From the balanced chemical equation, we can see that 1 mole of BX3 reacts with 3 moles of H2O to produce 1 mole of B(OH)3 and 3 moles of HX.

Therefore, the total number of moles of reactants involved in the reaction is 1 (from BX3) + 3 (from H2O) = 4 moles.

Knowee AI · Accepted Answer

The reaction for the preparation of boric acid from tri halide of boron and water is as follows:

BX3 + 3H2O → B(OH)3 + 3HX

Where BX3 is the tri halide of boron (Boron Trichloride, Boron Tribromide, etc.) and HX is a halogen acid (Hydrochloric acid, Hydrobromic acid, etc.)

From the balanced chemical equation, we can see that 1 mole of BX3 reacts with 3 moles of H2O to produce 1 mole of B(OH)3 and 3 moles of HX.

Therefore, the total number of moles of reactants involved in the reaction is 1 (from BX3) + 3 (from H2O) = 4 moles.

Total number of mole of reactants involved in the reaction for the preparation of boric acid from tri halide of boron and water is