Question

3 LiOH + H3PO4 ⟶ Li3PO4 + 3 H2O       SHOW WORK

A. How many grams of Li3PO4 are produced when 8.45 grams of LiOH are reacted with 9.53 grams of H3PO4?

B. What is the limiting reactant?

C. If you experimentally produce 7.15 grams of Li3PO4, calculate the percent yield.

Answer 1

A)

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

mass(LiOH)= 8.45 g

number of mol of LiOH,

n = mass of LiOH/molar mass of LiOH

=(8.45 g)/(23.976 g/mol)

= 0.3524 mol

Molar mass of H3PO4,

MM = 3*MM(H) + 1*MM(P) + 4*MM(O)

= 3*1.008 + 1*30.97 + 4*16.0

= 97.994 g/mol

mass(H3PO4)= 9.53 g

number of mol of H3PO4,

n = mass of H3PO4/molar mass of H3PO4

=(9.53 g)/(97.994 g/mol)

= 9.725*10^-2 mol

Balanced chemical equation is:

3 LiOH + H3PO4 ---> Li3PO4 + 3 H2O

3 mol of LiOH reacts with 1 mol of H3PO4

for 0.3524 mol of LiOH, 0.1175 mol of H3PO4 is required

But we have 0.0973 mol of H3PO4

so, H3PO4 is limiting reagent

we will use H3PO4 in further calculation

Molar mass of Li3PO4,

MM = 3*MM(Li) + 1*MM(P) + 4*MM(O)

= 3*6.968 + 1*30.97 + 4*16.0

= 115.874 g/mol

According to balanced equation

mol of Li3PO4 formed = (1/1)* moles of H3PO4

= (1/1)*0.0973

= 0.0973 mol

mass of Li3PO4 = number of mol * molar mass

= 9.725*10^-2*1.159*10^2

= 11.27 g

Answer: 11.27 g

B)

H3PO4 is limiting reagent

C)

% yield = actual mass*100/theoretical mass

= 7.15*100/11.27

= 63.45%

Answer: 63.45%