) How many grams of nitrogen is required to completely react with 54g of magnesium
to form magnesium nitride.
Answers
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Answer:
I think this was a right answer......
Explanation:
As per the equation, let us calculate the mole ratio.
N
2
+
3
H
2
→
2
N
H
3
As per the equation one mole of nitrogen reacts with 1 mol of hydrogen.
In terms of mass. 28.01 g of nitrogen needs 3 mol of hydrogen or 6.048 g of hydrogen.
We can set up the ratio;
28.01 g of
l
N
2
needs
6.048 g of
l
H
2
1 g of
l
N
2
needs
6.048
28.01
g of
l
H
2
50.0 g of
l
N
2
needs
6.048
×
50.0
28.01
l
g of
l
H
2
=
10.80 g of
l
H
2
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