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jolli1 [7]
3 years ago
12

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)

Chemistry
1 answer:
Blizzard [7]3 years ago
8 0
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
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An aqueous solution of calcium hydroxide is standardized by titration with a 0.112 M solution of hydrobromic acid. If 15.2 mL of
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0.0457 M

Explanation:

The reaction that takes place is:

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First we<u> calculate how many moles of acid reacted</u>, using the <em>HBr solution's concentration and volume</em>:

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  • 0.112 M * 12.4 mL = 1.389 mmol HBr

Now we <u>convert HBr moles to Ca(OH)₂ moles</u>, using the stoichiometric ratio:

  • 1.389 mmol HBr * \frac{1mmolCa(OH)_{2}}{2mmolHBr} = 0.6944 mmol Ca(OH)₂

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