Question
Question: How many moles of KOH are required to neutralize 196 grams of sulphuric acid? (Given \({{H}_{2}}S{{O...
How many moles of KOH are required to neutralize 196 grams of sulphuric acid? (Given H2SO4 = 98 amu)?
(A) 1.1
(B) 1.5
(C) 2.0
(D) 4.0
Solution
We know that sulphuric acid is a strong acid and potassium hydroxide is a strong base. Addition of a base to the acid makes the acid to reach a pH value of 7. If the solution reaches pH 7, then the reaction is called a neutralization reaction.
Complete step by step solution:
- In the question, it is given that to find the number of moles of potassium hydroxide required to neutralize 196 grams of sulphuric acid.
- The molecular formula of sulphuric acid is H2SO4.
- First, we have to write a balanced equation to get the information about the reaction.
- The reaction of sulphuric acid with potassium hydroxide is as follows.
2KOH+H2SO4→K2SO4+2H2O
- In the above reaction 2 moles of potassium hydroxide reacts with one mole of sulphuric acid and forms potassium sulphate and 2 moles of water as the products.
- Given that Molecular weight of sulphuric acid is 98 amu.
2moles2KOH+98gH2SO4→K2SO4+2H2O
- As per the above equation 2 moles of potassium hydroxide reacts with 98 grams of sulphuric acid.
- In the question it is asked to find the number of moles of potassium hydroxide required to react with 196 grams of sulphuric acid.
- Then,