A chemist has a beaker containing lead nitrate, Pb(NO3)2, dissolved in water. The chemist adds solution containing sodium iodide, NaI, and a bright yellow precipitate is formed. The chemist chemist continues to add NaI until no further yellow precipitate is formed. The chemist filters the precipitate until there is no more yellow precipitate in it. The chemist filters the precipitate, dries it in the oven, and it has a mass of 1.43 g.
A) write a balanced equation to explain what happened. Hint: compoinds with sodium ions are soluable.
B) determine the mass of Pb(NO3)2 that was dissolved
Can anyone walk me through this question??
A) write a balanced equation to explain what happened. Hint: compoinds with sodium ions are soluable.
B) determine the mass of Pb(NO3)2 that was dissolved
Can anyone walk me through this question??
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A) Pb(NO3)2 + 2 NaI -----> PbI2(ppt) + 2 NaNO3
B) calc. moles of PbI2 formed: 1.43 g PbI2 / 461 g/mole PbI2= 0.0031 moles PbI2
from the bal. rxn., 1 mole PbI2 is formed from 1 mole Pb(NO3)2
0.0031 moles x 331.2 g/mole Pb(NO3)2= 1.03 g Pb(NO3)2
B) calc. moles of PbI2 formed: 1.43 g PbI2 / 461 g/mole PbI2= 0.0031 moles PbI2
from the bal. rxn., 1 mole PbI2 is formed from 1 mole Pb(NO3)2
0.0031 moles x 331.2 g/mole Pb(NO3)2= 1.03 g Pb(NO3)2