Question

9. Potassium reacts aggressively with water, as shown in this equation: 2K + 2 H20 → 2 KOH + H2 If 0.2 grams of potassium are added to 15.0 grams of water, what mass of KOH can form in this reaction? (2 pts.)

Answer 1

form potassium reacts with water to KOH and liberate Hz ges. 2k + 2 H2O => 2 KOH + Hz. 2 role of k reacth usla a mole of H₂O to form a mole at koH and liberate Imole of H₂. 2 mole of k 2 mole of H₂O required. Molar mass of H₂O = 2x Homic maf& ot t't x Afшіс мая оно = 2x1+1*16= 18:9/mol. Molar massot k= 39 g/mol, 2 mole of k 2 mole of H₂O resuired (2x39ot k = (2x18) 30+ H2O resuired, 78 g of k 36 s of H₂O resuired, 0.2 gota 27 mate of Heo required = 0.2*36 78 = 0.0929

But 15.0 g of H₂O is present, 4-6 H2O is excess geagent. It is exces864 (15.0-0.092) 9 = 14,9089 potassium is consumed totally in the reaction are it is limiting reagent- 2 role of k → 2 mole of Koh formed, (2x39) 9 Of K = 2 x 56) 8 of Koly formed 78 got k > 112 g of kol formed 0.2 gat k - ?? mass of Koh formed = 0.2 x 112 78 = 0.2879 : 0.287 got kot can be formed