The combustion of ammonia in presence of excess oxygen yields NO2 and H2O.
The molar mass of ammonia is 17.02 g/mol
Therefore, moles of ammonia in 43.9 g
= 43.9 /17.02
= 2.579 moles
From the equation the mole ratio of ammonia to nitrogen iv oxide is 4:4
The molar mass of NO2 is 46 g/mol
The number of moles of NO2 is the same as that of ammonia since they have equal ratio,
= 2.579 moles
Therefore, mass of NO2
= 2.579 moles ×46
= 118.634 g
≈ 119 g
99.98%=0.9998
0.01%=0.0001
=(1)(0.9998)+2(0.0001)+3(0.0001)
=1.0003
Answer: Option (C) is the correct option.
Explanation:
When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.
For example,
....(1)
.........(2)
Cancelling the common species in both the equations as follows.
Therefore, on addition we get the equation as follows.
The number of moles in the resulting solution can be calculated by adding together the number of moles from the individual solution. That is calculated as follows,
total number of moles = (35 mL)(1 L/1000 mL)(1 mole/L) + (60 mL)(1 L/1000 mL)(0.6 moles / L) = 0.071 moles
The precipitation reaction is,
KBr + AgNO3 --> AgBr + KNO3
From the equation, each mole of the KBr will need one mole of AgNO3. Hence, 0.071 mole of AgNO3 will also be needed. Calculating for the mass,
mass of AgNO3 = (0.071 moles)(169.87 g / mol) = 12.06 g
ANSWER: 12.06 grams
