22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass of magnesium nitrate that will be produced.

22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass
of magnesium nitrate that will be produced.

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  1. Answer:

    9.82 g of Mg(NO₃)₂

    Explanation:

    Let’s determine the reaction:

    2AgNO₃  +  MgBr₂  → Mg(NO₃)₂  +  2AgBr

    2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

    We determine the moles of AgNO₃

    22.5 g . 1mol / 169.87g = 0.132 moles

    Ratio is 2:1.

    2 moles of silver nitrate can produce 1 mol of magnesium nitrate

    Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

    We convert moles to mass:

    0.0662 mol . 148.3 g/ mol = 9.82 g

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