How many grams of KClO₃ are needed to produce of 2.56 grams of O₂? 2KClO₃(s)→ 2KCl(s)+ 3O₂(g)

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First, we need to understand the chemical equation given, which is in a coded format. Decoding it, we get: 2KClO3 → 2KCl + 3O2. This equation tells us that 2 moles of potassium chlorate (KClO3) decompose to produce 2 moles of potassium chloride (KCl) and 3 moles of oxygen gas (O2). To find out how many grams of KClO3 are needed to produce 2.56 grams of O2, we first need to calculate the molar mass of O2, which is 32 g/mol (16 g/mol for each oxygen atom). Therefore, 2.56 grams of O2 is 2.56/32 = 0.08 moles of O2.According to the balanced chemical equation, 3 moles of O2 are produced from 2 moles of KClO3. Therefore, 0.08 moles of O2 would be produced from (2/3) * 0.08 = 0.0533 moles of KClO3.The molar mass of KClO3 is approximately 122.55 g/mol (39.1 for K, 35.45 for Cl, and 48 for O3). Therefore, 0.0533 moles of KClO3 would weigh 0.0533 * 122.55 = 6.53 grams.

How Many Grams of KClO3 Are Needed to Produce of 2.56

How Many Grams of KClO₃ Are Needed to Produce of 2.56