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10 June, 18:39

how many grams of water are produced when 15 grams of magnesium hydroxide are reacted with sulfuric acid

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  1. 10 June, 18:50
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    9.26 g of water

    Explanation:

    We are given;

    15 grams of magnesium hydroxide;

    Required to calculate the mass of water produced;

    We will use the following steps;

    Step 1: Write the balanced equation for the reaction

    Magnesium hydroxide reacts with sulfuric acid to produce water and magnesium sulfate;

    Therefore, the equation of the reaction is;

    Mg (OH) ₂ (aq) + H₂SO₄ (aq) → MgSO₄ (aq) + 2H₂O (l)

    Step 2: Determine the number of moles of Mg (OH) ₂ used

    Number of moles = Mass : Molar mass

    Molar mass of Mg (OH) ₂ = 58.320 g/mol

    Therefore;

    Moles of Mg (OH) ₂ = 15 g : 58.320 g/mol

    = 0.257 moles

    Step 3: Calculate the moles of water produced

    From the equation 1 mole of Mg (OH) ₂ reacts to produce 2 moles of water;

    Therefore; Moles of water = moles of Mg (OH) ₂ * 2

    Moles of water = 0.257 moles * 2

    = 0.514 moles

    Step 4: Calculate the mass of water produced;

    Mass = Moles * Molar mass

    Molar mass of water = 18.02 g/mol

    Therefore;

    Mass of water = 0.514 moles * 18.02 g/mol

    = 9.26 g

    Therefore, 9.26 g of water will be produced
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