Question

if you have 3.50 moles of hydrogen and 5.00moles of nitrogen to produce ammonia, witch element is the reactant in excess? 3H2+N2=2NH3

Answer 1

Nitrogen is the N and that is what is the element is the reactant in excess.

Answer 2

Answer : The $N_2$ element is the reactant in excess.

Solution :  Given,

Moles of $N_2$ = 5 moles

Moles of $H_2$ = 3.50 moles

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 moles of $H_2$ react with 1 mole of $N_2$

So, 3.5 moles of $H_2$ react with $\frac{3.5}{3}=1.16$ moles of $N_2$

The excess of $N_2$ = 5 - 1.16 = 3.84 moles

That means in the given balanced reaction, $H_2$ is a limiting reagent because it limits the formation of products and $N_2$ is an excess reagent.

Hence, the $N_2$ element is the reactant in excess.