Question

How many grams of potassium chlorate decompose to potassium chloride and 719 mL of O2 at 128°C and 685 torr?

2 KClO3(s) → 2 KCl(s) + 3 O2 (g)

*please explain

Answer #1

**Given:**

**P = 685.0 torr**

**= (685.0/760) atm**

**= 0.901316 atm**

**V = 719.0 mL**

**= (719.0/1000) L**

**= 0.719 L**

**T = 128.0 oC**

**= (128.0+273) K**

**= 401 K**

**find number of moles using:**

**P * V = n*R*T**

**0.901316 atm * 0.719 L = n * 0.08206 atm.L/mol.K * 401
K**

**n = 1.968*10^-2 mol**

**This is number of moles of O2 formed**

**2 KClO3(s) → 2 KCl(s) + 3 O2 (g)**

**From balanced reaction above,**

**moles of KClO3 reacted = (2/3)*moles of O2
formed**

**= (2/3)*1.968*10^-2 mol**

**= 0.01312 mol**

**Molar mass of KClO3,**

**MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)**

**= 1*39.1 + 1*35.45 + 3*16.0**

**= 122.55 g/mol**

**mass of KClO3,**

**m = number of mol * molar mass**

**= 1.312*10^-2 mol * 122.55 g/mol**

**= 1.61 g**

**Answer: 1.61 g**

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