Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 grams of lithium and 25.0 grams of bromine are present at the beginning of the reaction. limiting reactant

Bromine is the limiting reactant

Explanation:

First of all let's generate a balanced equation for the reaction

2Li + Br2 —> 2LiBr

Molar Mass of Li = 7g/mol

Molar Mass of Br2 = 2x80 = 160g/mol

From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:

Number of mole of Li = 25/7 = 3.6moles

Number of mole of Br2 = 25/160 = 0.156mol.

To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

For the experimental yield:

Li : Br2 = 3.6/ 0.156 = 23 : 1

For the theoretical yield:

Li : Br = 2 : 1

From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.