Answer:
Hydrogen, 0,2 mol
Explanation:
[tex]V_{M}[/tex] = 22, 4 l/mol
V ([tex]N_{2}[/tex]) = 6,72 [tex]dm^{2}[/tex] = 6,72 l
V ([tex]H_{2}[/tex]) = 6,72 [tex]dm^{2}[/tex] = 6,72 l
The formula is:
n = [tex]\frac{V}{V_{M} }[/tex]
n ([tex]N_{2}[/tex]) = 6,72l /22,4 l/mol = 0,3 mol
n ([tex]H_{2}[/tex]) = 6,72l /22,4 l/mol = 0,3 mol (the same)
But from the equation we get a proportion that 1 mol ([tex]N_{2}[/tex]) - 3 mol ([tex]H_{2}[/tex])
[tex]\frac{0,3}{1} > \frac{0,3}{3}[/tex] (we have more hydrogen than is need for the reaction)
So there is excess of [tex]H_{2}[/tex]
We only need [tex]\frac{0,3}{3}[/tex] = 0,1 mol ([tex]H_{2}[/tex])
0,3 mol (hydrogen quantity we have) - 0,1 mol (hydrogen quantity we need for the reaction) = 0,2 mol (excess)
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