Pb(NO3)2 + 2 NaI PbI2 + 2 NaNO3 a. If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? b. What is the limiting reactant? c. How many grams of lead (II) iodide is formed
Pb(NO3)2 + 2 NaI PbI2 + 2 NaNO3 a. If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? b. What is the limiting reactant? c. How many grams of lead (II) iodide is formed
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Chapter3: Calculations With Chemical Formulas And Equaitons
Section: Chapter Questions
Problem 3.19QP: You react nitrogen and hydrogen in a container to produce ammonia, NH3(g). The following figure...
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When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed.
Pb(NO3)2 + 2 NaI PbI2 + 2 NaNO3
a. If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium
nitrate can be formed?
b. What is the limiting reactant?
c. How many grams of lead (II) iodide is formed?
d. How much of the excess reagent will be left over?
e. If 6 grams of sodium nitrate are formed in the reaction, what is the percent yield of this reaction?
*Please focus on sub parts D and E
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