Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3
a.) How many moles of lead (II) nitrate are needed to react completely with 10 moles of sodium iodide?
1 Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3 The ratio of lead nitrate to sodium iodide is 1/2
x/10 = 1/2 x = 5 moles
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