Question

25.0 g of Li2O reacts in 500mL of water to produce 22.3 g LiOH. What are the limiting reactant, the theoretical and percent yields?
Li2O + H2O --> 2 LiOH

Answer 1

1)

Molar mass of Li2O,

MM = 2*MM(Li) + 1*MM(O)

= 2*6.968 + 1*16.0

= 29.936 g/mol

mass(Li2O)= 25.0 g

number of mol of Li2O,

n = mass of Li2O/molar mass of Li2O

=(25.0 g)/(29.936 g/mol)

= 0.8351 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

density of water = 1 g/mL

mass(H2O)= density * volume = 1 g/ml * 500 mL = 500 g

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(500 g)/(18.016 g/mol)

= 27.75 mol

Balanced chemical equation is:

Li2O + H2O ---> 2 LiOH +

1 mol of Li2O reacts with 1 mol of H2O

for 0.8351 mol of Li2O, 0.8351 mol of H2O is required

But we have 27.7531 mol of H2O

so, Li2O is limiting reagent

Answer: Li2O is limiting reagent

b2)

we will use Li2O in further calculation

Molar mass of LiOH,

MM = 1*MM(Li) + 1*MM(O) + 1*MM(H)

= 1*6.968 + 1*16.0 + 1*1.008

= 23.976 g/mol

According to balanced equation

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

= (2/1)*0.8351

= 1.6702 mol

mass of LiOH = number of mol * molar mass

= 1.67*23.98

= 40.05 g

Answer: Theoretical yield = 40.05 g

3)

% yield = actual mass*100/theoretical mass

= 22.3*100/40.05

= 55.7%

Answer: percent yield = 55.7 %