Question

Calculate the moles of K₂SO₄ of a solution made by mixing 550.0mL 0.35M H₂SO₄ and 375.0mL 0.55M KOH. Assume 100% yield for both reactions.

                        H₂SO₄(aq) + 2 KOH(aq) -> 2 H₂O(l) + K₂SO₄(aq)  

                        H₂SO₄(aq) + 2 H₂O -> 2 H₃O⁺(aq) + SO₄^2-(aq)

Answer 1

Molarity(M)=number of moles of solute/Volume of solution in L

Number of moles of solute=MolarityxVolume of solution in L

1L=1000 mL

So 1mL=10^-3 L

So Number of moles of solute=Molarityx(10^-3 L/mL)Volume of solution in mL

So number of moles of KOH=0.55Mx(10^-3 L/mL)x375mL=0.20625=0.206

Number of moles of H₂SO₄=0.35Mx(10^-3 L/mL)x550mL=0.192

H₂SO₄ + KOH$\rightarrow$K₂SO₄ + H₂O

So for one mole of H₂SO₄ reacting with 1 mole KOH, 1 mole K₂SO₄ is formed

Here H₂SO₄ is the limiting reactant as it is present in less amount, so for 0.192 moles of H₂SO_4, 0.192 moles KOH are consumed and 0.192 moles K₂SO₄ are formed.