For the reaction 2k(s)+br2(l)→2kbr(s) calculate how many grams of the product form when 21.4 g of br2 completely reacts. assume that there is more than enough of the other reactant.
the balanced equation for the above reaction is as follows; 2K + Br₂ --> 2KBr Stoichiometry of Br₂ to KBr is 1:2 we have been told that K is present which means that Br₂ is the limiting reactant. Limiting reactant is the reactant that is fully consumed in the reaction and amount of product formed depends on amount of limiting reactant present. Number of Br moles reacted - 21.4 g / 160 g/mol = 0.134 mol According to stoichiometry, 1 mol of Br₂ forms 2 mol of KBr therefore 0.134 mol of Br₂ forms 2x0.134 = 0.268 mol of KBr mass of KBr formed - 0.268 mol x 119 g/mol = 31.9 g of KBr a mass of 31.9 g of KBr is formed
