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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?

 Mar 3, 2022
edited by Alan  Mar 3, 2022
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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

 Mar 3, 2022

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