To produced 340 of NH3
How many gram nitrogen requried
Answers
Answer:
280
Explanation:
ratio by mass 14:3
hence 340/16=20
14×20=280
Answer:
The required mass of nitrogen is 12.34 g.
Explanation:
The balanced equation is:
0.5
N
2
+
1.5
H
2
→
NH
3
...
...
...
...
...
...
...
...
...
...
...
.
(
1
)
The molecular mass of ammonia,
NH
3
, is equal to 14.01 g + 3(1.008 g), or 17.03 g.
Moles of ammonia
=
mass in grams
molecular mass
=
15.00 g
17.03 g/mol
=
15.00
17.03
mol
Therefore, according to equation
(
1
)
,
Moles of nitrogen
=
0.5
(
15.00
17.03
)
mol
...
...
...
...
(
2
)
The number of moles of nitrogen is calculated by the following equation:
Moles of N
2
=
mass of nitrogen
molar mass of nitrogen
Thus,
Mass of nitrogen
=
moles of N
2
⋅
molar mass of N
2
...
(
3
)
Therefore, from equations
(
2
)
and
(
3
)
, we have:
Mass of nitrogen
=
0.5
(
15.00
17.03
)
mol
×
28.02 g/mol
=
12.34 g