If the percent yield for the following reaction is 65.0%,how many grams of KClO₃ are needed to produce 42.0 g of O₂? 2 KClO₃(s)→ 2 KCl(s)+ 3 O₂(g) A)69.7 g B)107 g C)165 g D)371 g

Question

Quizplus · Accepted Answer

To find the grams of KClO3 needed, first calculate the theoretical yield of O2. The molar mass of O2 is 32.00 g/mol, so 42.0 g of O2 is 42.0 g / 32.00 g/mol = 1.3125 mol O2. From the balanced equation, 3 mol of O2 are produced from 2 mol of KClO3. Therefore, 1.3125 mol O2 corresponds to (2/3) * 1.3125 mol = 0.875 mol KClO3. The molar mass of KClO3 is 122.55 g/mol, so 0.875 mol KClO3 is 0.875 mol * 122.55 g/mol = 107.23 g KClO3. Since the percent yield is 65.0%, the actual amount needed is 107.23 g / 0.65 = 165 g.

If the Percent Yield for the Following Reaction Is 65