N(NaI):n(PbI2)=2:1
so n(PbI2)=3.0mol
m(PbI2)=3.0*453=1359g.
so n(PbI2)=3.0mol
m(PbI2)=3.0*453=1359g.
How Many Grams Of PbI2 Are Produced From 6.0 Mol NaI In The Following Reaction? Pb(NO3)2 + 2 NaI ® 2 NaNO3 + PbI2