Question

N2+3H2=>>2NH3. If 6.72dm³ of each of Nitrogen and hydrogen (measured at S.T.P.) are made to produced Ammonium. Which of the reacted will be in excess and by how many moles?

Answer 1

Answer:

Hydrogen, 0,2 mol

Explanation:

$V_{M}$ = 22, 4 l/mol

V ($N_{2}$) = 6,72 $dm^{2}$ = 6,72 l

V ($H_{2}$) = 6,72 $dm^{2}$ = 6,72 l

The formula is:

n = $\frac{V}{V_{M} }$

n  ($N_{2}$) = 6,72l /22,4 l/mol = 0,3 mol

n  ($H_{2}$) = 6,72l /22,4 l/mol = 0,3 mol (the same)

But from the equation we get a proportion that 1 mol  ($N_{2}$) - 3 mol ($H_{2}$)

$\frac{0,3}{1} > \frac{0,3}{3}$ (we have more hydrogen than is need for the reaction)

So there is excess of $H_{2}$

We only need  $\frac{0,3}{3}$ = 0,1 mol ($H_{2}$)

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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