Answer:
The answer to your question is a) H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
b) Molarity = 0.127
Explanation:
Data
[KOH] = 0.133 M
volume of KOH = 28.2 ml
[H₂SO₄] = ?
volume of H₂SO₄ = 15 ml
1) Balanced chemical reaction
H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
Reactants Elements Products
2 K 2
1 S 1
4 H 4
6 O 6
2) Calculate the molar mass of H₂SO₄ and KOH
H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4) = 2 + 32 + 64 = 98 g
KOH = (39 x 1) + (16 x 1) + (1 x 1) = 39 + 16 + 1 = 56 g
- Calculate the number of moles of KOH
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = 0.133 x 28.2/1000
= 0.0038
- Calculate the moles of H₂SO₄
1 mol of H₂SO₄ ------------------ 2 moles of KOH
x ------------------ 0.0038 moles
x = (0.0038 x 1)/2
x = 0.0019 moles of H₂SO₄
-Calculate the molarity of H₂SO₄
Molarity = 0.0019/0.015
Molarity = 0.127
