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26 March, 08:21

Calculate the mass of silver bromide produced from 22.5 g of silver nitrate.

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  1. 26 March, 08:28
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    Equation is as follow,

    2 AgNO ₃ + MgBr ₂ → 2 AgBr + Mg (NO ₃) ₂

    According to eq.

    339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr

    So,

    22.5 g AgNO₃ will produce = X g of AgBr

    Solving for X,

    X = (22.5 g * 375.54 g) : 339.74 g

    X = 24.87 g of AgBr
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