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12 November, 06:41

For the reaction Pb (NO3) 2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 246.2 g of potassium iodide?

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  1. 12 November, 08:26
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    M (KI) = 246.2 g

    M (KI) = 166.0 g/mol

    n (KI) = m (KI) / M (KI)

    n (KI) = 246.2/166.0≈1.48 mol

    2 mol KI - 1 mol PbI₂

    1.48 mol KI - x mol PbI₂

    x=1.48*1/2=0.74 mol
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