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5 October, 02:52

2KHCO3 (s)  K2CO3 (s) + CO2 (g) + H2O (l) How many moles of potassium carbonate will be produced if 454 g of potassium hydrogen carbonate are heated? A) 2.27 mol B) 3.29 mol C) 11.4 mol D) 227 mol E) 4.54 mol

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Answers (2)
  1. 5 October, 03:15
    0
    A

    Explanation:

    gram mol

    2KHCO3 K2CO3

    200 1

    454 X

    X=454:200=2.27g
  2. 5 October, 03:18
    0
    2.27 moles potassium carbonate will be produced. (Option A is correct.)

    Explanation:

    Step 1: Data given

    Mass of potassium hydrogen carbonate (KHCO3) = 454 grams

    Molar mass KHCO3 = 100.1 g/mol

    Step 2: The balanced equation

    2KHCO3 (s) → K2CO3 (s) + CO2 (g) + H2O (l)

    Step 3: Calculate moles KHCO3

    Moles KHCO3 = mass KHCO3 / molar mass KHCO3

    Moles KHCO3 = 454 grams / 100.1 g/mol

    Moles KHCO3 = 4.54 moles

    Step 4: Calculate moles of K2CO3

    For 2 moles KHCO3 we'll have 1 mol K2CO3, 1 mol CO2 and 1 mol H2O

    For 4.54 moles KHCO3 we'll have 4.54/2 = 2.27 moles K2CO3

    2.27 moles potassium carbonate will be produced. (Option A is correct.)
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