if 50.0 g of KCl reacts with 50.0 g of O2 to produce KClO3 according to the equation, how many grams of KClO3 will be formed?

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  1. 2K Cl + 3O2 --> 2KClO3
    KCl = 39+ 35.5 = 74.5 g/mol
    O = 16 g/mol so O2 is 32 g/mol
    I need 3 mols of O2 for every 2 mols of KCl
    50 g KCl = 50/74.5 = .671 mols of KCl
    so need .671 * 3/2 = 1.007 mols of O2 = 32 g
    SO I have MORE O2 than I need, just use the .671 mols of KCl, limiting reagent
    I get 1 mol KClO3 for every mol of KCl
    so I get .671 mols of KClO3
    74.5 + 3*16 = 122.5 grams/mol of KClO3
    .671 * 122.5 = 2.2 grams

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