How much heat energy is required to boil 34.9 g of ammonia, NH,? The molar heat...

Question

Question

How much heat energy is required to boil 34.9 g of ammonia, NH,? The molar heat of vaporization of ammonia is 23.4 kJ/mol. q=

When heated, KCIO, decomposes into KCl and 02. 2 KCIO, — 2 KCl + 302 If this reaction produced 54.0 g KCl, how many grams of

What volume of 0.215 M KOH, V, can be neutralized with 54.0 mL of 1.56 M HNO,? Vs ml

Answer 1

Answer #1

1. The molar heat of vaporization of ammonia is 23.4 kJ/mol. This means that 23.4 kJ of heat is required to boil one mole of ammonia. So to boil

, the heat energy required is

2. The balanced chemical equation for the decomposition of KClO₃ is

mass of KCl formed = 54.0 g

moles of KCl formed

The balanced equation shows that the two products form in a molar ratio of 2:3. Therefore, moles of oxygen gas formed

In terms of mass, oxygen formed is

3. Nitric acid, HNO₃ and potassium hydroxide, KOH, react in a 1:1 mole ratio to produce aqueous potassium nitrate, KNO₃, and water.

moles of HNO₃ in 54.0 ml of 1.56 M solution

Therefore, to neutralize 0.08424 mol of HNO₃, 0.08424 mol of KOH is needed. Therefore, the volume of KOH needed can be calculated as

Answer 2

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