If 348 grams of Mg(OH)2 reacts with excess HCl (this is CL chlorine, not C and I), how many moles of MgCl2 (this is CL chlorine, not C and I

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If 348 grams of Mg(OH)2 reacts with excess HCl (this is CL chlorine, not C and I), how many moles of MgCl2 (this is CL chlorine, not C and I) are produced?

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Diễm Kiều 1 month 2021-08-16T05:25:38+00:00 1 Answers 0 views 0

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    2021-08-16T05:26:57+00:00

    Answer:

    moles MgCl₂ = 5.97 moles

    Explanation:

    First, let0s write the overall reaction that is taking place here:

    Mg(OH)₂ + HCl ———-> MgCl₂ + H₂O

    Now, we need to balance this reaction:

    Mg(OH)₂ + 2HCl ———-> MgCl₂ + 2H₂O

    Now, the exercise already stated that the HCl is in excess, so we can assume the magnesium hydroxide is the limiting reactant. If this is true, then, to get the moles of MgCl₂ produced, all we need to do is calculate the moles of Mg(OH)₂ that reacted, using the following expression:

    moles = mass / MM

    The atomic weights for Mg, O and H are:

    Mg: 24.305 g/mol;  H: 1 g/mol;  O = 15.999 g/mol

    Let’s determine the molar mass of Mg(OH)₂:

    MM = 24.305 + (2 * 15.999) + (2 * 1) = 58.303 g/mol

    Now, let’s determine the moles:

    moles = 348 g / 58.303 g/mol = 5.97 moles of Mg(OH)₂

    As we can see in the overall reaction, there is a mole ration between Mg(OH)₂ and MgCl₂ of 1:1, that’s why we can conclude that the moles of Mg(OH)₂ would be the same moles of MgCl₂:

    moles Mg(OH)₂ = moles MgCl₂ = 5.97 moles

    Hope this helps

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