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11 July, 22:34

K + H2O - -> KOH + H2

1. How many grams of KOH are produced if 224 g of H2O are used?

2. How many moles of K are used if 20 moles of H2O react?

3. How many grams of H2O are used if 3678 grams of H2 are produced?

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  1. 12 July, 00:44
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    1. 696.9g of KOH

    2. 20moles of K

    3. 66204g of H2O

    Explanation:

    First, we need to generate a balanced equation for the reaction. This is illustrated below:

    2K + 2H2O - > 2KOH + H2

    Next, we'll find the molar mass and mass of the reactants and products as shown below:

    Molar Mass of K = 39g/mol

    Mass of K from the balanced equation = 2x39 = 78g

    Molar Mass of H2O = (2x1) + 16 = 18g/mol

    Mass of H2O from the balanced equation = 18 x 2 = 36g

    Molar Mass of KOH = 39 + 16 + 1 = 56g/mol

    Mass of KOH from the balanced equation = 2 x 56 = 112g

    Molar Mass of H2 = 2x1 = 2g/mol

    1. 2K + 2H2O - > 2KOH + H2

    From the balanced equation,

    36g of H20 produced 112g of KOH.

    Therefore, 224g of H2O will produce = (224 x 112) / 36 = 696.9g of KOH

    2. 2K + 2H2O - > 2KOH + H2

    From the balanced equation,

    2moles of K required 2moles of H2O.

    Therefore, Xmol of K will require 20moles of H2O i. e

    Xmol of K = (20 x 2) / 2 = 20moles

    3. 2K + 2H2O - > 2KOH + H2

    From the balanced equation,

    36g of H2O produced 2g of H2.

    Therefore,

    Xg of H2O will produce 3678g of H2 i. e

    Xg of H2O = (3678 x 36) / 2 = 66204g
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