Potassium nitrate problem for ap chem!

Question

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

a) write balanced equation

b) what mass of KNO3 would be needed to produce 18.4 liters of O2, measeured at 775 mmHg and 15 degrees C?

c) what mass of KNO2 would also be produced?

Answer 1

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

b) This is not STP therefore use ideal gas law to find the number of mols of O2.

PV = nRT

rearrange: n = PV /RT

n = (775mmHg / 760mmHg/atm) (18.4L) / (0.08206 L atm/mol K) ( 15 + 273.15K)

n= 0.79mols of O2

0.79 mols X 2(Stoichiometry) = 1.58mols of KNO3

molar mass of KNO3 = 101g/mol

101g/mol X 1.58mols = 159.6 g of KNO3

c) The molar mass of KNO2 is 85g/mol

85g/mol X 1.58mols = 134.3g of KNO2

Answer 2

2KNO3 => 2KNO2 + O2

775 mmHg = 1.02 atm

15 C = 288 K

18.4 L O2 @STP ~ 0.821 mol O2

.821 mol O2 x ( 2 mol KNO3 / 1 mol O2) x (101.1032 g KNO3 / 1 mol KNO3) ~ 166.01 g KNO3 needed to produce 18.4 L O2

Since there is a 1:1 ratio between KNO3 and KNO2, you can not do any math and know that is you use 166.01 g KNO3, you will end up with 166.01 g KNO2.