Question

A 15­g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride. 2Li + F2 → 2LiF After the reaction is complete, what will be present? (20. g of lithium fluoride and 9.5 g of lithium)

Answer 1

Molar mass of Li = 6.968 g/mol


mass(Li)= 15.0 g

use:
number of mol of Li,
n = mass of Li/molar mass of Li
=(15.0 g)/(6.968 g/mol)
= 2.153 mol

Molar mass of F2 = 38 g/mol


mass(F2)= 15.0 g

use:
number of mol of F2,
n = mass of F2/molar mass of F2
=(15.0 g)/(38 g/mol)
= 0.3947 mol
Balanced chemical equation is:
2 Li + F2 ---> 2 LiF


2 mol of Li reacts with 1 mol of F2
for 2.153 mol of Li, 1.076 mol of F2 is required
But we have 0.3947 mol of F2

so, F2 is limiting reagent
we will use F2 in further calculation


Molar mass of LiF,
MM = 1*MM(Li) + 1*MM(F)
= 1*6.968 + 1*19.0
= 25.968 g/mol

According to balanced equation
mol of LiF formed = (2/1)* moles of F2
= (2/1)*0.3947
= 0.7895 mol


use:
mass of LiF = number of mol * molar mass
= 0.7895*25.97
= 20.5 g
According to balanced equation
mol of Li reacted = (2/1)* moles of F2
= (2/1)*0.3947
= 0.7895 mol
mol of Li remaining = mol initially present - mol reacted
mol of Li remaining = 2.153 - 0.7895
mol of Li remaining = 1.363 mol


Molar mass of Li = 6.968 g/mol

use:
mass of Li,
m = number of mol * molar mass
= 1.363 mol * 6.968 g/mol
= 9.5 g
Answer:
mass of LiF formed = 20.5 g
mass of Li remaining = 9.5 g
These two are present at the end