If an aqueous solution of sulfuric acid (H2SO4) is neutralized by 3.20 mol of sodium hydroxide (NaOH), how many moles of acid were originally present in the solution?
If completely neutralized, the balanced equation is
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
The equation tell you that 2 moles NaOH = 1 mole H2SO4 so 3.2 moles NaOH must equal ?? moles H2SO4.
To determine the number of moles of acid originally present in the solution, we can use the concept of stoichiometry.
The balanced chemical equation for the neutralization reaction between sulfuric acid and sodium hydroxide is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Given that 3.20 mol of sodium hydroxide is used, we can use the stoichiometry to find the number of moles of sulfuric acid.
Number of moles of sulfuric acid = (Number of moles of sodium hydroxide) / 2
Number of moles of sulfuric acid = 3.20 mol / 2
Number of moles of sulfuric acid = 1.60 mol
Therefore, there were originally 1.60 moles of sulfuric acid present in the solution.
To solve this question, we need to use the concept of stoichiometry and balanced chemical equations. Here's how you can approach it step by step:
1. Write a balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH):
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
2. Identify the stoichiometric ratio between sulfuric acid and sodium hydroxide from the balanced chemical equation. In this case, it is 1:2. This means that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
3. Determine the number of moles of sodium hydroxide given in the question: 3.20 mol.
4. Since the stoichiometric ratio is 1:2, the number of moles of sulfuric acid would be half of the number of moles of sodium hydroxide used. Therefore, divide the number of moles of sodium hydroxide by 2: 3.20 mol / 2 = 1.60 mol.
Hence, there were originally 1.60 moles of sulfuric acid in the aqueous solution.