Answer :
Explanation:
M
r
N
H
3
=
17
;
M
r
O
2
=
32
;
M
r
N
O
=
26
;
M
r
H
2
O
=
18
Using the equation:
Number of Moles = Mass / Mr
Number of Moles
N
H
3
=
175
17
=
10.3
Number of Moles
O
2
=
310
32
=
9.69
O
2
is the limiting reagent because the least moles of this are used in the reaction.
From the balanced equation, we can see that
O
2
and
N
O
are in a 5:4 ratio
5
4
=
0.8
9.69
×
0.8
=
7.75
7.75 moles of
N
O
is the maximum that can be produced from 310g of
O
2
. To convert this to mass we use the same equation as in the first step.
Mass = Moles x Mr
7.75
×
26
=
202
g
202g is the theoretical yield of
N
O
To calculate the percentage yield, you just divide the actual yield by the theoretical yield and multiply by 100:
197
202
×
100
=
98
%
The limiting reactant for the reaction between 10 moles of NH₃ and 30 moles of NO is NH₃.
The balanced equation for the reaction is given below:
4NH₃ + 6NO → 5N₂ + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted with 6 moles of NO.
With the above information, we can obtain the limiting reactant as follow:
From the balanced equation above,
4 moles of NH₃ reacted with 6 moles of NO.
Therefore,
10 moles of NH₃ will react with = [tex]\frac{10 * 6}{4}\\\\[/tex] = 15 moles of NO.
From the calculation made above, we can see clearly that only 15 moles of NO out of 30 moles reacted completely with 10 moles of NH₃.
Therefore, NH₃ is the limiting reactant and NO is the excess reactant.
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