1. Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.50-

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(ClO3)2+ NaI -------------------------------> PbI + Na(ClO3)2

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. CO2 is produced from burning of sugar only as it is a carbon containing compound

C12H22O11 + 12 O2 ----> 12 C02+ 11 H2O

1 mol of sugar gives 12 mol of CO2, i.e. 1:12 ratio

Molar mass of CO2= 44 gm/mol

Moles of CO2= 2.40/44= 0.054 moles

This will be produced from 0.054/12= 0.0045 moles of sugar

1 mol of sugar= 342 g/mol

0.0045 mol of sugar= 342*0.0045= 1.53 gm

So, salt present= 5.50-1.53= 3.97 gm

2. Molar mass of CuNO3= 125.5 gm

1 L of 0.180 M contains 0.180 moles of CuNO3

mass of CuNO3= 0.180*125.5= 22.59 gm

2.31 g of CuNO3 correspond to 2.31/22.59= 0.10 L solution

3. equation and values are not matching. Please correct the question.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.