Question

if 1.00g of oxygen is allowed to react with 3.00g of lithium, what mas of lithium oxide can form?

Answer 1

Molar mass of Li = 6.968 g/mol

mass of Li = 3.0 g

we have below equation to be used:

number of mol of Li,

n = mass of Li/molar mass of Li

=(3.0 g)/(6.968 g/mol)

= 0.4305 mol

Molar mass of O2 = 32 g/mol

mass of O2 = 1.0 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(1.0 g)/(32 g/mol)

= 3.125*10^-2 mol

we have the Balanced chemical equation as:

4 Li + O2 ---> 2 Li2O

4 mol of Li reacts with 1 mol of O2

for 0.4305 mol of Li, 0.1076 mol of O2 is required

But we have 3.125*10^-2 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of Li2O = 2*MM(Li) + 1*MM(O)

= 2*6.968 + 1*16.0

= 29.936 g/mol

From balanced chemical reaction, we see that

when 1 mol of O2 reacts, 2 mol of Li2O is formed

mol of Li2O formed = (2/1)* moles of O2

= (2/1)*3.125*10^-2

= 6.25*10^-2 mol

we have below equation to be used:

mass of Li2O = number of mol * molar mass

= 6.25*10^-2*29.94

= 1.87 g

Answer: 1.87 g