Question

For the following reaction, 61.9 grams of barium hydroxide are allowed to react with 40.2 grams of sulfuric acid . barium hydroxide(aq) + sulfuric acid(aq) barium sulfate(s) + water(l) What is the maximum amount of barium sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1)

Molar mass of Ba(OH)2,

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

= 1*137.3 + 2*16.0 + 2*1.008

= 171.316 g/mol

mass(Ba(OH)2)= 61.9 g

use:

number of mol of Ba(OH)2,

n = mass of Ba(OH)2/molar mass of Ba(OH)2

=(61.9 g)/(1.713*10^2 g/mol)

= 0.3613 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 40.2 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(40.2 g)/(98.09 g/mol)

= 0.4098 mol

Balanced chemical equation is:

Ba(OH)2 + H2SO4 ---> BaSO4 + 2 H2O

1 mol of Ba(OH)2 reacts with 1 mol of H2SO4

for 0.3613 mol of Ba(OH)2, 0.3613 mol of H2SO4 is required

But we have 0.4098 mol of H2SO4

so, Ba(OH)2 is limiting reagent

we will use Ba(OH)2 in further calculation

Molar mass of BaSO4,

MM = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

According to balanced equation

mol of BaSO4 formed = (1/1)* moles of Ba(OH)2

= (1/1)*0.3613

= 0.3613 mol

use:

mass of BaSO4 = number of mol * molar mass

= 0.3613*2.334*10^2

= 84.32 g

Answer: 84.3 g

2)

Ba(OH)2 is limiting reagent

3)

According to balanced equation

mol of H2SO4 reacted = (1/1)* moles of Ba(OH)2

= (1/1)*0.3613

= 0.3613 mol

mol of H2SO4 remaining = mol initially present - mol reacted

mol of H2SO4 remaining = 0.4098 - 0.3613

mol of H2SO4 remaining = 4.852*10^-2 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

use:

mass of H2SO4,

m = number of mol * molar mass

= 4.852*10^-2 mol * 98.09 g/mol

= 4.76 g

Answer: 4.76 g