Question

Two molecules of lithium are combined with 225 grams of bromine to form two molecules of lithium bromide. If you end up with 690 grams of solid lithium bromide, how much lithium did you start with.

Answer 1

The amount of Li present to start the reaction is 55.18g

Explanation:

2Li + Br₂ → 2LiBr

Molecular weight of Br₂ = 159.808 g/mol

Mass of Br₂ present = 225 g

Moles of Br₂ present during the reaction = 225 / 159.808

                                                                m = 1.4

Molecular weight of LiBr = 86.845 g/mol

Mass of LiBr formed = 690 g

moles of LiBr produced = 690 / 86.845

                               m(LiBr) = 7.95

According to the balanced equation, 2 molecules of Li reacts to for 2 molecule of LiBr

So, 7.95 moles of LiBr would require 7.95 moles of Li

The molecular weight of Li is 6.941 g/mol

Thus, the amount of lithium present to start the reaction is

$moles = \frac{given weight}{molecular weight} \\\\7.95 = \frac{w}{6.941} \\\\w = 55.18g$

Therefore, the amount of Li present to start the reaction is 55.18g