If 1.27 moles of bromine gas (Br2) react with excess phosphorus (P), how many moles of phosphorus tribromide (PBr3) will be produced?

0.85 mole of PBr3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Br2 + 2P —> 2PBr3

From the balanced equation above,

3 moles of Br2 reacted to produce 2 moles of PBr3.

Therefore, 1.27 moles of Br2 will react to produce = (1.27 x 2)/ 3 = 0.85 mole of PBr3.

Therefore, 0.85 mole of PBr3 is produced by the reaction.

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