[Solved] Write the balanced neutralization reaction that occurs - General Chemistry I (CHE 111) - Studocu

Balanced Neutralization Reaction

The balanced neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) is as follows:

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

This reaction shows that one mole of sulfuric acid reacts with two moles of potassium hydroxide to produce one mole of potassium sulfate and two moles of water.

Calculation of Remaining Sulfuric Acid Concentration

To calculate the remaining concentration of sulfuric acid after neutralization, we first need to calculate the number of moles of H2SO4 and KOH.

• Moles of H2SO4 = Volume (L) x Molarity (M)
• Moles of KOH = Volume (L) x Molarity (M)

Let's calculate:

• Moles of H2SO4 = 0.350 L x 0.440 M = 0.154 moles
• Moles of KOH = 0.300 L x 0.280 M = 0.084 moles

From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, all the KOH will react with half its moles of H2SO4.

• Moles of H2SO4 reacted = 0.084 moles / 2 = 0.042 moles

Subtracting the moles of H2SO4 reacted from the initial moles of H2SO4 gives the remaining moles of H2SO4.

• Remaining moles of H2SO4 = 0.154 moles - 0.042 moles = 0.112 moles

Finally, we can calculate the remaining concentration of H2SO4 by dividing the remaining moles of H2SO4 by the total volume of the solution (0.350 L + 0.300 L = 0.650 L).

• Remaining concentration of H2SO4 = 0.112 moles / 0.650 L = 0.172 M

So, the concentration of sulfuric acid that remains after neutralization is 0.172 M.