Suppose 0.950 L of 0.410 MH, SO, is mixed with 0.900 L of 0.240 M KOH. What concentration of sulfuric acid remains

after neutralization?

The remaining concentration of H2SO4 is 0.152 M

Explanation:

Step 1: Data given

Volume of H2SO4 = 0.950 L

Molarity H2SO4 = 0.410 M

Volume of KOH = 0.900 L

Molarity of KOH = 0.240 M

Step 2: The balanced equation

H2SO4 + 2KOH → K2SO4 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles H2SO4 = 0.410 M * 0.950 L

Moles H2SO4 = 0.3895 moles

Moles KOH = 0.240 M * 0.900L

Moles KOH = 0.216 moles

Step 4: Calculate the limiting reactant

For 1 mol H2SO4 we need 2 moles KOH to produce 1 mol K2SO4 and 2 moles H2O

The limiting reactant is KOH. It will completely be consumed (0.216 moles).

H2SO4 is in excess. There will react 0.216/2 = 0.108 moles. There will remain 0.3895 moles - 0.108 moles = 0.2815 moles

Step 5: Calculate the concentration of H2SO4 remaining

[H2SO4] = moles / volume

[H2SO4] = 0.2815 moles / 1.85 L

[H2SO4]= 0.152 M

The remaining concentration of H2SO4 is 0.152 M

thats how its supposed to be