Question #22136
When Lead (2) Nitrate reacts with sodium iodide, sodium nitrate and lead (2) iodide are formed. if i start with 25.2 grams of lead (2) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
Expert's answer
To solve we should do two calculations. Atfirst,we will determine the quantity of sodium nitrate that can be formed with25 grams of lead (II) nitrate, assuming that there’s plenty of sodium iodide
present to react with it
n of Pb(NO3)2 = 25.2/331 = 0.076
n of NaI = 15.0/150 = 0,1
0.076 n
Pb(NO3)2 + 2NaI = 2NaNO3 + PbI2
1 2
n NaNO3 = 0.076*2/1 = 0.152mol
m = 85 * 0.152 = 12.92g
present to react with it
n of Pb(NO3)2 = 25.2/331 = 0.076
n of NaI = 15.0/150 = 0,1
0.076 n
Pb(NO3)2 + 2NaI = 2NaNO3 + PbI2
1 2
n NaNO3 = 0.076*2/1 = 0.152mol
m = 85 * 0.152 = 12.92g
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#340153 on Dec 2023
#340153 on Dec 2023