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  • one year ago

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) 2.72 g KClO3 0.361 g KClO3 83.6 kg KClO3 22.5 mg KClO3

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  1. Photon336
    • one year ago
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    2KClO3(s) --> 2KCl(s) + 3O2 assuming the reaction goes to completion: given 2.72grams of KCLO3 covert this to moles: so it would be 2.72g x (1mol/123g) notice grams cancel out and you're left with moles. then you get roughly 0.0221mol KClO3. then you multiply the number of moles by the molar ratio of KClO3 to Oxygen to find the number of moles of oxygen produced. 0.0221 mol KClO3 x molar ratio (3O2/2KClO3) (notice KCLO3 in numerator and denominator so it cancels out leaving you with the moles of oxygen when you multiply the two quantities, so you get 0.0221 x(3/2) = 0.0332 mol of oxygen. 0.0332 mol x (32g O2/1mol) = 1.1g O2 you would follow a similar process for the other steps ensuring that you convert milligrams to grams before you do any calculations.

  2. Photon336
    • one year ago
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    note this reaction is a decomposition going from A-->B+C. Also note that they told you this reaction has gone to completion, so you don't have to find a limiting reagent. if you did however, you would find which compound is limiting FIRST and then calculate the number of moles/grams of products produced by using the limiting reagent.

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