Question

For the reaction 2K(s)+Br2(l)→2KBr(s) calculate how many grams of the product form when 20.4 g of Br2 completely reacts. Assume that there is more than enough of the other reactant.

Answer 1

2K + Br₂ --> 2KBr
Ratio of moles of Br₂ to KBr is 1:2

It is given that K is present in excess which means that Br₂ is the limiting reactant. Limiting reactant is the reactant that is fully consumed in the reaction and product amount depends upon limiting reagent.
Number of Br moles reacted = mass of Br2/molar mass of Br2 =20.4 g / 159.8g/mol = 0.127 mol
According to stoichiometry,
1 mol of Br₂ forms 2 mol of KBr
therefore 0.127 mol of Br₂ forms 2x0.127= 0.254 mol of KBr

mass of KBr formed =number of moles of KBr×molar mass of KBr =0.254mol x 119 g/mol = 30.22g of KBr
a mass of 30.22 g of KBr is formed .

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