SHOW WORK for points 5. Ammonium carbonate decomposes upon heating according to

Question

SHOW WORK for points

5. Ammonium carbonate decomposes upon heating according to the following equation: (NH4) 2CO3 (s) 2NH3 (g) + CO2 (g) + H2O (g). Calculate the total volume of gas produced at 22 °C and 1.02 atm by complete decomposition of 11.83 g of ammonium carbonate.

6. A scuba diver creates a spherical bubble with r = 2.5 cm at a depth of 30.0 m where the total pressure (including atmospheric pressure) is 4.00 atm. What is the radius of the bubble when it reaches the surface of the water? Assume atmospheric pressure to be 1.00 atm, and the temperature 298 K.

7. Which of the noble gasses in Group VIIIA of the periodic table has a density of 3.7493 g/L at 0 °C and 1.00 atm?

8. Calculate the volume of CO2 gas collected at 23 °C and 0.991 atm that is prepared by reacting 10.0g of calcium carbonate with excess acid: CaCO3 (aq) + 2H+(aq) Ca2+(aq) + CO2 (g) + H2O(l)

9. 3.57 g of a KCl-KClO3 mixture is decomposed by heating and produces 119 mL O2 gas (at 22.4 °C and 738 mmHg). What is the percent of KClO3 in the mixture? 2KClO3 (s) 2KCl(s) + 3O2 (g)

Explanation / Answer

5. Given 1mole ammonium carbonate can produce 2mole ammonia, so 0.123mole(11.83/96.09) can produce 0.246mole of ammonia.

Now PV = nRT

         V = (0.246*0.0821*295)/1.02 = 5.846 litres   ( R = Molar gas constant = 0.0821 lit-atm K-1 mol-1

6.Sorry I don't have knowledge about it

7. From PV = nRT

   P = [(w/M)RT] / V ( since n = weight(w)/Molecular weight(M) )

      = (d RT) / M

M = dRT/P = (3.7493*0.0821*273)/1   = 84.034

This molecular belongs to Krypton (Kr)

8.Given 1mole Calcium carbonate can produce 1mole CO2, so 0.1mole(10/100) can produce 0.1mole of CO2.

Now PV = nRT

         V = (0.1*0.0821*296)/0.991 = 2.452 litres   ( R = Molar gas constant = 0.0821 lit-atm K-1 mol-1)

9. 3.57g mixure can produce 119ml O2(at 22.4 °C and 738 mmHg = 0.971atm) So

      from PV = nRT

             n = PV / RT = (0.971*0.119) / (0.0821*295.4) = 0.00476 moles of O2 is produced

Given 2 mole KClO3 can produce 3mole O2, So 0.00476 mole O2 was produced by 0.003173mole KClO3

n = w/M

w = 0.003173* 122.55 = 0.3888grams

% of KClO3 in the mixture = (0.3888/3.57)*100 = 10.89%

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