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 [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
[tex]{\text {Number of moles of bromine}}=\frac{16.7g}{80g/mol}=0.21moles[/tex]
Thus [tex]Br_2[/tex] acts as limiting reagent as it limits the formation of product as potassium is in excess.
[tex]2K(s)+Br_2(l)\rightarrrow 2KBr(s)[/tex]
According to stoichiometry:
1 mole of [tex]Br_2[/tex] reacts to give 2 moles of [tex]KBr[/tex]
Thus 0.21 moles of [tex]Br_2[/tex] will react to give=[tex]\frac{2}{1}\times 0.21=0.42[/tex] moles of [tex]KBr[/tex]
Mass of [tex]KBr=moles\times {\text {Molar mass}}=0.42moles\times 119g/mol=49.7grams[/tex]
49.7 grams of the product will be formed.
