150g of magnesium nitride on hydrolysis gave 22.4 L of NH3 at STP the % yeild reaction is (mg3+N2+6H2O 2NH3+3mg(OH)2)
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Answer:
Mg
3
N
2
(
s
)
+
6
H
2
O
(
l
)
→
3
Mg
(
OH
)
2
(
a
q
)
+
2
NH
3
(
a
q
)
Explanation:
Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.
For example, we have
Mg
3
N
2
on the left side and on the right side on the original equation we see magnesium as
Mg
(
OH
)
2
.
On the left side, there are 3 moles of
Mg
. Now making sure we have exactly 3 moles of
Mg
on the right side, we add a
3
in front of the compound
Mg
(
OH
)
2
to make it
3
Mg
(
OH
)
2
. Do same for all.
Explanation:
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