Given the balanced neutralization equation: H2SO4+2KOH→K2SO4+2H2O
How many moles of potassium hydroxide (KOH) are required to neutralize 4.5 mol of sulfuric acid (H2SO4)? Assume that the sulfuric acid completely dissociates in water.
2 mols of KOH = 1 mol of H2SO2
4.5 mols of H2SO4 = 9 mols of KOH
right. Go the head of the class.
To determine the number of moles of potassium hydroxide (KOH) required to fully neutralize 4.5 moles of sulfuric acid (H2SO4), we need to use the balanced neutralization equation provided:
H2SO4 + 2KOH → K2SO4 + 2H2O
According to this equation, for every 1 mole of sulfuric acid, we need 2 moles of potassium hydroxide. Therefore, the ratio is 2 moles of KOH to 1 mole of H2SO4.
To find the number of moles of KOH needed, we can use the following calculation:
Number of moles of KOH = (Number of moles of H2SO4) * (Ratio of KOH to H2SO4)
Number of moles of KOH = 4.5 mol * (2 mol KOH / 1 mol H2SO4)
Number of moles of KOH = 9 mol
Therefore, to neutralize 4.5 moles of sulfuric acid (H2SO4), we need 9 moles of potassium hydroxide (KOH).