Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
To balance the following chemical equation, make a tally or a count of each of the atoms on both sides of the reaction, and make sure that those atoms are equal on both the reactant and product side.
AL2O3 + HCl => ALCl3 + H2O
Left side. Right side
AL = 2. AL = 1
O = 3 Cl = 3
H = 1. H = 2
Cl = 1. O = 1
First balance the metal atoms, aluminum, then hydrogen and then oxygen.
Balanced equation :
AL2O3 + 6HCl => 2ALCl3 + 3H2O.
Left side. Right side
AL = 2 AL = 2
O = 3 Cl = 6
H = 6 H = 6
Cl = 6 O = 3.
