Chem question

Question

1.)Potassium nitrate decomposes when heated, forming potassium nitrite and oxygen gas.

           a.) Write a balanced chemical equation for this reaction

           b.) What mass of KNO₃ would be needed to produce 18.4                     L of oxygen gas, measured at 775 mm Hg and 15°C?                c.) What mass of KNO₂ would also be produced?

2.) When solutions of potassium iodide and lead (II) nitrate are combined, the products are potassium nitrate and bright yellow lead (II) iodide.

           a.) Write a balanced equation for this reaction, including                     (aq) and (s).

           b.) Calculate the mass of precipitate produced when 50.0                    mL of 0.45 M potassium iodide solution and 75 ml of                     0.55 M lead (II) nitrate solution are mixed.

           c.) Calculate the volume of 0.50 M potassium iodide                              required to react completely with 50.0 ml of 0.50 M                          lead (II) nitrate.

 

Answer 1

1.

a) 4KNO3 ---> 4KNO2 + 2O2

b) moles of O2(g) = n = PV/RT (from the ideal gas law) and n = (775mm)(18.4L)/(62.36L-mm/deg-mol)(288deg)

n = 0.794 moles of O2 gas.  From the balanced equation it can be seen that 1 mole O2(g) comes from 2 moles KNO3 (the mole ratio is 4 moles KNO3 : 2 moles O2).  Thus, moles KNO3 needed = 2 x 0.794 = 1.588 moles KNO3 needed.

Mass KNO3 = 1.588 moles x 101.1 gram/mole = 160.5 grams needed

c) Mass of KNO2 produced?  From the balanced equation, one can see that the mole ratio of KNO2 : KNO3 is 1:1, i.e. 4 moles KNO3 produces 4 moles KNO2.  Thus, moles KNO2 produced from  1.588 moles KNO3 = 1.588 moles KNO2.

 

2.) 

a)  2KI(aq) + Pb(NO3)2(aq) ---> 2KNO3(aq) + PbI2(s)

b)  moles KI used = (0.05 L)(0.45 mole/L) = 0.0225 moles KI

moles Pb(NO3) used = (0.075 L)(0.55 mol/L) = 0.04125 moles Pb(NO3)2

What is limiting in this reaction?  From the balanced equation, it takes twice as many moles of KI as it does Pb(NO3). So, for 0.04125 moles Pb(NO3), one needs 2 x that or 0.0825 moles KI, but there is only 0.0225 moles. Thus, KI is limiting.  What does this mean.  It means that the most precipitate (PbI2) that can be made is limited by the moles of KI present.  2 moles KI produces 1 mole PbI2 (see balanced equation).

Moles of PbI2 produced = 1/2 the moles of KI used = 1/2 x 0.0225 moles = 0.01125 moles PbI2 produced

mass of PbI2 = 0.01125 moles x 461 g/mole = 5.19 grams

c)  moles of KNO3 = 0.05 L x 0.5 mol/L = 0.025 moles KNO3

moles KI needed to react with 0.025 moles KNO3 = 2 x 0.025 = 0.05 moles KI needed (see mole ratio of 2:1 in balanced equation)

Volume KI needed:  0.5 mol/L (X L) = 0.05 moles

X = 0.1 liters needed

 

Answer 2

   KNO₃  -->   KNO₂  +   O₂

 

NO₃⁺ is the nitrate ion.

 

NO₂⁺ is the nitrite ion.

 

First count the number of oxygen atoms on the right side. You have 4: 2 with potassium nitrite and 2 in a molecule of oxygen.

Now it is all trial and error, but after some guess and check, you might discover the 3 on the left is going to be part of a common multiple with 4.  The lowest common multiple is 12.  How do you get 12 Oxygen atoms?

 

4 KNO₃  --> 4 KNO₂  +  2 O₂