Question

Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide are mixed with 0.278 moles of oxygen gas. What is the formula for the limiting reagent? What is the maximum amount of nitrogen dioxide that can be produced?

Answer 1

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

       According to question

                                2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

$'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278$

Since 'NO' is the limiting reagent according to this ratio.

According to equation

         2 moles NO reacts to form 2 moles NO₂

So,  0.1855 moles NO give  = 0.1855 moles of NO₂

            Mass of 1 mole NO₂ = 46 g/mole

            Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g