2 If you used 71 g of K and 200 g of Br2, how much Br2 would you have leftover? 2 K + Br2 --> 2 KB

Answer 1

Stoichiometry for each reactant:

(71g K) * (1 mole K /39.10g K) * (2 mole KBr /2 mole K) = 1.8₁₆ mole KBr.

(200g Br₂) * (1 mole Br₂ /159.8g Br₂) * (2 mole KBr /1 mole Br₂) = 2.5₀₃ mole KBr.

Remaining (Δmole = 0.68₇ mole KBr) * (1 mole Br₂ /2 mole KBr) * (159.8g Br₂ /1 mole Br₂) = 55g Br₂.