Calculate how many grams of the product form when 16.7 g of liquid bromine reacts with solid potassium. assume that there is more than enough of the solid potassium. 2 k(s) + br2(l) → 2 kbr(s)

49.7 grams of the product will be formed.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number of particles.

To calculate the moles, we use the equation:

Thus acts as limiting reagent as it limits the formation of product as potassium is in excess.

According to stoichiometry:

1 mole of reacts to give 2 moles of

Thus 0.21 moles of will react to give= moles of

Mass of

49.7 grams of the product will be formed.