I know all of the answers but please explain how they are found :) (1) The combu
ID: 905323 • Letter: I
Question
I know all of the answers but please explain how they are found :)
(1) The combustion of diborane (B2H6) generates a lot of energy. Balance the following equation for the reaction of diborane with oxygen
B2H6 (g) + O2 (g) B2O3 (s) + H2O (g)
What is the stoichiometric coefficient for oxygen?
I know the answer is 3
(2) How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.11 g/mol.
4 KNO3 (s) 2 K2O (s) + 2 N2 (g) + 5 O2 (g)
I know the answer is .724 mol O2
(3) A 14.01 g sample of N2 reacts with 3.02 g of H2 to form ammonia (NH3). If ammonia is the only product, what mass of ammonia is formed?
I know the answer is 17.03 g
Explanation / Answer
1)
B2H6 (g) + O2 (g) B2O3 (s) + H2O (g)
You need to balance it
Balance B... its already balanced (2/2)
Balance H... (6/2)
B2H6 (g) + O2 (g) B2O3 (s) + 3H2O (g)
balance O.. (2, 6)
B2H6 (g) + 3O2 (g) B2O3 (s) + 3H2O (g)
Therefore, the stoichiometric factor must be 3
2)
4 KNO3 (s) 2 K2O (s) + 2 N2 (g) + 5 O2 (g)
moles of O2 when m = 58.6 g of KNO3 decompose MW = 101.11
calculate moles of KNO3
mol = mass/MW = 58.6 / 101.11 = 0.5796 mol
equation is already balanced. relate KNO3 and O2
4 moles of KNO3 --> 5 mol of O2
ratio must be then 5/4 mol = 5/4*0.5796 mol of O2 will be produced = 0.7245 mol of O2
3)
m = 14.01 g of N2
MW = 28
mol = 14/28 = 0.5 mol of N2
m = 3.02 H2
MW = 2
mol = 3/2= 1.5 mol
The reaction
N2 + H2 --> NH3
Balanced equation
N2+ 3H2 --> 2NH3
Ratio is
1 mol of N2 --> 2 mol of NH3
since we have 0.5 mol of N2... then we have 2*0.5 = 1 mol of NH3
MW of NH3 = 17 g/mol
mass = mol*MW = 1*17 = 17 g of NH3