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2 years ago

13

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

1 answer:

marta [7]2 years ago

7 0

Answer:

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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