Question

For the following reaction, 4.39 grams of nitrogen gas are allowed to react with 8.77 grams of oxygen gas. nitrogen (g) + oxygen (g nitrogen monoxide (g) What is the maximum amount of nitrogen monoxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

Balanced reaction is

N₂ (g) + O₂ (g) $\to$   2 NO (g)

Mole ratio of nitrogen and oxygen = 1:1

Molar mass of nitrogen = 28 g/mol

Molar mass of oxygen = 32 g/mol

So, 28 g Nitrogen reacts with 32 g Oxygen.

Or, 1 g Nitrogen reacts with  _{$\frac{32}{28}$} g of Oxygen.

Or, 4.39 g Nitrogen reacts with (4.39 × _{$\frac{32}{28}$} ) = 5.017 g oxygen.

But mass of oxygen taken = 8.77 g

Mass of oxygen is excess so limiting reagent is nitrogen

Excess amount of oxygen

= (8.77 - 5.017) g

= 3.753 g.