Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
Answer:1) Volume of
required is 55.98 mL.
2) 0.62577 grams of
is produced.
Explanation:

1) Molarity of 
Volume of 
Molarity of 
Volume of 


According to reaction, 1 mole of
reacts with 3 mole of
, then, 0.0041985 moles of
will react with:
moles of
that is 0.0125955 moles.


Volume of
required is 55.98 mL.
2)

Number of moles of
According to reaction, 3 moles of
gives 1 mole of
, then 0.004485 moles of
will give:
moles of
that is 0.001495 moles.
Mass of
=
Moles of
× Molar Mass of 
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of
is produced.
The answer for the following problem is mentioned below.
- <u><em>Therefore the final moles of the gas is 14.2 × </em></u>
<u><em> moles.</em></u>
Explanation:
Given:
Initial volume (
) = 230 ml
Final volume (
) = 860 ml
Initial moles (
) = 3.8 ×
moles
To find:
Final moles (
)
We know;
According to the ideal gas equation;
P × V = n × R × T
where;
P represents the pressure of the gas
V represents the volume of the gas
n represents the no of the moles of the gas
R represents the universal gas constant
T represents the temperature of the gas
So;
V ∝ n
= 
where,
(
) represents the initial volume of the gas
(
) represents the final volume of the gas
(
) represents the initial moles of the gas
(
) represents the final moles of the gas
Substituting the above values;
= 
= 14.2 ×
moles
<u><em>Therefore the final moles of the gas is 14.2 × </em></u>
<u><em> moles.</em></u>