Question

H₂SO₄(aq) + KOH(aq) ---> H₂O(l) + K₂SO₄(aq)   

Determine what mass of water is produced when a beaker containing 100.0g H₂SO₄ dissolved in 250 mL of water is added to a larger beaker containing 100.0g KOH dissolved in 225 mL of water. Determine the mass amounts of each substance (other than water) present in the large beaker when the reaction is complete.

Answer 1

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)= 100.0 g

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(100.0 g)/(98.086 g/mol)

= 1.02 mol

Molar mass of KOH,

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

= 1*39.1 + 1*16.0 + 1*1.008

= 56.108 g/mol

mass(KOH)= 100.0 g

number of mol of KOH,

n = mass of KOH/molar mass of KOH

=(100.0 g)/(56.108 g/mol)

= 1.782 mol

Balanced chemical equation is:

H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

1 mol of H2SO4 reacts with 2 mol of KOH

for 1.019513 mol of H2SO4, 2.039027 mol of KOH is required

But we have 1.782277 mol of KOH

so, KOH is limiting reagent

we will use KOH in further calculation

Molar mass of K2SO4,

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

= 2*39.1 + 1*32.07 + 4*16.0

= 174.27 g/mol

According to balanced equation

mol of K2SO4 formed = (1/2)* moles of KOH

= (1/2)*1.782277

= 0.891139 mol

mass of K2SO4 = number of mol * molar mass

= 0.8911*1.743*10^2

= 155 g

According to balanced equation

mol of H2SO4 formed = (1/2)* moles of KOH

= (1/2)*1.782277

= 0.891139 mol

mol of H2SO4 remaining = mol initially present - mol reacted

mol of H2SO4 remaining = 1.019513 - 0.891139

mol of H2SO4 remaining = 0.128375 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 of H2SO4,

m = number of mol * molar mass

= 0.1284 mol * 98.086 g/mol

= 12.g g

Answer:

mass of H2SO4 = 12.6 g

mass of KOH = 0

mass of K2SO4 = 155 g