1. Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.50-
ID: 881068 • Letter: 1
Question
1. Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.50-g sample is burned, and 2.40 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?
2. If 2.31 g of CuNO3 is dissolved in water to make a 0.180 M solution, what is the volume of the solution?
3. Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction:
Pb(NO3)2+ 2NH4I -------------------------------> PbI2 + 2NH4NO3
What volume of a 0.570 M NH4I solution is required to react with 965 mL of a 0.280 M Pb(NO3)2 solution?
4. Rank these species by their ability to act as an oxidizing agent: Cu2+, Au3+, Ca2+, and Fe2+
5. Write the balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI. Include phases.
6. What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.750 L of 0.190 M NaI? Assume the reaction goes to completion.
7. If a solution containing 17.30 g of mercury(II) nitrate is allowed to react completely with a solution containing 5.102 g of sodium sulfate, how many grams of solid precipitate will be formed?
How many grams of the reactant in excess will remain after the reaction?
8. The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation.
Given that it requires 38.9 mL of 0.210 M Na2S2O3(aq) to titrate a 30.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.
Explanation / Answer
1. C12H22O11 + 12 O2 = 12 CO2 + 11 H2O
1 mole sugar = 12 mole CO2
No OF Moles of Co2 = 2.4/32 = 0.075 mole
No of moles of sugar burned = 0.075/12 = 0.00625 mole
mass of sugra = 0.00625*342 = 2.1375 grams
salt present = 5.5-2.1375 = 3.3625 grams
2. wt of CuNO3 = 2.31 , molarity = 0.18 M
v = ?
0.18 = (2.31/125.55)*1000/v in ml
Volume = 102.217 ml
3. Pb(NO3)2+ 2NH4I(aq) ---> PbI2(s) + 2NH4NO3(aq)
No of moles of Pb(NO3)2 = 0.28*0.965 = 0.2702 mole
No of moles of NH4I = 2*0.2702 = 0.5404
volume of NH4I = 0.5404/0.57 = 0.948 L
4. Au3+ > Cu+2>Fe+2> Ca+2
------------------------> decreases.
5. Pb(ClO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaClO3 (aq)
6. No of moles of NaI = 0.75*0.19 = 0.1425 mole
No of moles of Pb(ClO3)2 reacted = 0.1425/2 = 0.07125 mole
No of moles of PbI2 precipitated = 0.07125 mole
mass of precipitate = 0.07125*461.01 = 32.847 grams
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