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7 March, 00:29

How many grams of lead (II) iodide are produced from 6.000 moles of NaI according to the balanced equation: Pb (NO3) 2 + 2 NaI à 2 NaNO3 + PbI2

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  1. 7 March, 00:46
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    mass PbI₂ formed = 1383 grams

    Explanation:

    Pb (NO₃) ₂ + 2NaI = > 2NaNO₃ + PbI₂ (s)

    6 mol NaI = > 1/2 (6 mol) PbI₂ = 3 mol PbI₂ x 461.01 g/mol = 1383.03 grams ≅ 1383 grams (4 sig. figs.)
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