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13 March, 03:18

According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide? 3KOH (aq) + H3PO4 (aq) K3PO4 (aq) + 3H2O (l)

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  1. 13 March, 07:13
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    There is 37.36 grams of K3PO4 produced

    Explanation:

    Step 1: Data given

    Mass of H3PO4 = 29.6 grams

    KOH is in excess

    Molar mass of KOH = 56.11 g/mol

    Molar mass of H3PO4 = 97.99 g/mol

    Step 2: The balanced equation

    3KOH (aq) + H3PO4 (aq) ⇔ K3PO4 (aq) + 3H2O (l)

    Step 3: Calculate mass of KOH

    Mass KOH = mass KOH / molar mass KOH

    Mass KOH = 29.6 grams / 56.11 g/mol

    Mass KOH = 0.528 moles

    Step 4: Calculate moles of K3PO4

    Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

    For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4

    Step 5: Calculate mass of K3PO4

    Mass K3PO4 = moles K3PO4 * molar mass K3PO4

    Mass K3PO4 = 0.176 moles * 212.27 g/mol

    Mass K3PO4 = 37.36 grams

    There is 37.36 grams of K3PO4 produced
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