In: Chemistry
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
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