How much potassium chloride to make 6L of oxygen gas ?

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SUMMARY

This discussion focuses on calculating the amount of potassium chloride (KCl) required to produce 6 liters of oxygen gas (O2) using the decomposition reaction of potassium chlorate (KClO3). The reaction is represented by the equation KClO3 = KCl + 3/2 O2. To determine the moles of oxygen gas at 24.0°C and 0.950 atm, the Ideal Gas Law is applied, leading to the conclusion that approximately 0.245 moles of O2 are needed, which corresponds to a specific weight of KClO3 for the reaction.

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1. Calculates the weight of potassium chloride needed to prepare 6.00L of oxygen gas at the temperature of 24.0 C and 0,950 atm.
2. The equation is : KClO3 = KCl + 3/2O2
3. A hint please ?
 
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