Na₂O va K₂O lardan iborat 12,5 g aralashma 100 g 16,8% li KOH eritmasida eritildi. Hosil bo‘lgan eritmani to‘liq neytrallash uchun 87,6 ml (ρ=1,25 g/ml) 20% li HCl eritmasi sarflandi. Dastlabki aralashmadagi kaliy oksidining massasini (g) toping.

  1. 9,4
  2. 4,7
  3. 2,35
  4. 7,05

📝 Detailed explanation

n(HCl) = 87,6·1,25·0,2/36,5 = 0,6 mol; n(KOH dastlab) = 16,8/56 = 0,3 mol. Oksidlardan kelgan ishqor = 0,6 − 0,3 = 0,3 mol OH⁻. Na₂O (x mol) va K₂O (y mol): 62x + 94y = 12,5 va 2x + 2y = 0,3. Yechib: x = 0,05; y = 0,1. m(K₂O) = 0,1·94 = 9,4 g.
8 / 30 (27%)
Question 8 of 30
← Previous → Next Space — Show answer

🎯 Take a test on the topic «Aprobatsiya KIMYO 2026»

This topic contains 30 questions. Test your knowledge!

🔐 Log in to take the test

📚 All tests by subject «Kimyo»

View all collections and topics in Kimyo

Go to subject →

Report an error

Answers are unavailable

This test collection is closed. To get full access, please purchase 🌟 premium

Contact by email