Question

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)

Answer 1

Answer: 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 $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

${\text {Number of moles of bromine}}=\frac{16.7g}{80g/mol}=0.21moles$  

Thus $Br_2$ acts as limiting reagent as it limits the formation of product as potassium is in excess.

$2K(s)+Br_2(l)\rightarrrow 2KBr(s)$

According to stoichiometry:

1 mole of $Br_2$ reacts to give 2 moles of $KBr$

Thus 0.21 moles of $Br_2$ will react to give=$\frac{2}{1}\times 0.21=0.42$ moles of $KBr$

Mass of $KBr=moles\times {\text {Molar mass}}=0.42moles\times 119g/mol=49.7grams$

49.7 grams of the product will be formed.