For the following reaction, 13.5 grams of nitrogen monoxide are allowed to react with 5.17 grams of oxygen gas .

Question

For the following reaction, 13.5 grams of nitrogen monoxide are allowed to react with 5.17 grams of oxygen gas .

nitrogen monoxide(g) + oxygen(g) ---> nitrogen dioxide(g)

What is the maximum mass of nitrogen dioxide that can be formed? _____ grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete? _____ grams

Answer 1

2NO(g) + O₂(g) ==> 2NO₂(g) ... balanced equation

13.5 g NO x 1 mol NO/30 g x 2 mol NO2/2 mol NO x 46 g NO2/mol = 20.7 g NO2 formed

5.17 g O2 x 1 mol O2/32 g x 2 mol NO2/1 mol O2 x 46 g NO2/mol = 14.9 g NO2 formed

Maximum mass NO2 formed = 149 g

Formula of limiting reagent = O2

Mass of excess reagent (NO) remaining is...

13.5 g NO x 1 mol NO/30 g = 0.45 moles initially present

5.17 g O2 x 1 mol/32 g x 2 mole NO used/mol O2 = 0.323 moles NO used up

mass NO remaining = 0.45 mol - 0.323 mol = 0.127 mol x 30 g/mol = 3.81 g NO remaining