Question

25. A 16.42-mL volume of 0.1327 M KMnO solution is needed to oxidize 2500 mL of a FeSO: solution in an acidic medium. What is the concentration of the Feso solution in molarity? The net ionic equation is 5Fe2+ MnOr - BH - Mn + 5Fe + 4H2O [8 pts] 26. A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate (KCIO.). Assuming complete decomposition, calculate the number of grams of Oz gas that can be obtained from 14 62 g KCIÓ. (The products are KCl and Ou). (8 pts] 2KC's) - 2KCs) + 30g)

Answer 1

25.

Mole ratio of KMnO₄ and Fe²⁺ = 1:5

Given ,

mmoles of KMnO₄ = molarity × volume

= ( 0.1327× 16.42)

= 2.178

So, mmoles of Fe²⁺ = moles of FeSO₄

= 5× mmoles of KMnO₄

= 5× 2.178

= 10.89.

Now,

Molarity of FeSO₄ × volume of FeSO₄ solution

= 10.89

Or, molarity of FeSO₄ × 25.00 = 10.89

Or, molarity of FeSO₄= (10.89/25) = 0.436 M

Therefore , molarity of FeSO₄ solution is 0.436 M

26.

Balanced reaction is

2KClO_{3 (s)}  $\rightarrow$ 2KCl (s) + 3O₂ (g)

Mole ratio of KClO₃ and O₂ = 2:3

Moles of KClO₃ = ( mass of KClO₃ / molar mass of KClO₃)

So, moles of KClO₃

= ( 14.62/122.5)

= 0.119

Hence, moles of Oxygen will form

= $\frac{3}{2}$ × Moles of KClO₃

= $\frac{3}{2}$ × 0.119

= 0.179

Now, mass of oxygen that can be obtained

= Moles × molar mass

= (0.179 × 32)

= 5.73 g.