The reaction between nitric oxide (NO) and oxygen to form nitrogen dioxide (NO2) is a key step in photochemical smog formation.
2 NO(g) + O2(g) 2 NO2(g)
(a) How many moles of NO2 are formed by the complete reaction of 0.404 mole of O2?
(b) How many grams of NO2 are formed by the complete reaction of 1.44 g of NO?
2NO(g) + O2(g) ==> 2NO2(g)
1. 0.404 mols NO x (2 mols NO2/1 mol O2) = ? mols NO2 formed.
2. mols NO = grams NO/molar mass NO = ?
Convert mols NO to mols NO2 with the coefficients like so.
? mols NO x (2 mols NO2/2 mols NO) = mols NO2 formed.
Then mols NO2 x molar mass NO2 = grams NO2 = ?
Post your work if you get stuck.
Answers
To determine the number of moles of NO2 formed by the complete reaction of a given amount of O2, we need to use stoichiometry. The balanced equation tells us that the ratio of moles of O2 to moles of NO2 is 1:2.
(a) How many moles of NO2 are formed by the complete reaction of 0.404 mole of O2?
Given: 0.404 mole of O2
From the balanced equation, we know that 1 mole of O2 reacts to form 2 moles of NO2.
So, using the equation:
Moles of NO2 = Moles of O2 × (2 moles of NO2 / 1 mole of O2)
Moles of NO2 = 0.404 mole × 2
Moles of NO2 = 0.808 mole
Therefore, 0.808 mole of NO2 are formed by the complete reaction of 0.404 mole of O2.
(b) How many grams of NO2 are formed by the complete reaction of 1.44 g of NO?
Given: 1.44 g of NO
First, we need to convert grams of NO to moles of NO using the molar mass of NO.
Molar mass of NO = (1 × atomic mass of N) + (1 × atomic mass of O)
= (1 × 14.01 g/mol) + (1 × 16.00 g/mol)
= 30.01 g/mol
Moles of NO = Mass of NO / Molar mass of NO
Moles of NO = 1.44 g / 30.01 g/mol
Moles of NO ≈ 0.048 mole
From the balanced equation, we know that 2 moles of NO react to form 2 moles of NO2.
So, using the equation:
Moles of NO2 = Moles of NO × (2 moles of NO2 / 2 moles of NO)
Moles of NO2 = 0.048 mole × 2
Moles of NO2 = 0.096 mole
Now we can convert moles of NO2 to grams using the molar mass of NO2.
Molar mass of NO2 = (1 × atomic mass of N) + (2 × atomic mass of O)
= (1 × 14.01 g/mol) + (2 × 16.00 g/mol)
= 46.01 g/mol
Mass of NO2 = Moles of NO2 × Molar mass of NO2
Mass of NO2 = 0.096 mole × 46.01 g/mol
Mass of NO2 ≈ 4.42 g
Therefore, 4.42 grams of NO2 are formed by the complete reaction of 1.44 g of NO.
To answer these questions, we'll use mole ratios from the balanced chemical equation and convert between moles and grams using molar masses.
(a) To find the number of moles of NO2 formed by the complete reaction of 0.404 mole of O2, we'll use the mole ratio between O2 and NO2.
From the balanced equation, we see that 1 mole of O2 reacts to form 2 moles of NO2.
So, using the mole ratio of O2 to NO2, we have:
0.404 mole O2 * (2 moles NO2 / 1 mole O2) = 0.808 moles NO2
Therefore, 0.808 moles of NO2 are formed by the complete reaction of 0.404 mole of O2.
(b) To find the grams of NO2 formed by the complete reaction of 1.44 g of NO, we'll use the mole ratio between NO and NO2.
First, we need to convert grams of NO to moles of NO.
Using the molar mass of NO (30.01 g/mol), we have:
1.44 g NO * (1 mole NO / 30.01 g NO) = 0.048 moles NO
Next, we'll use the mole ratio of NO to NO2 from the balanced equation, which is 2 moles NO2 for every 2 moles NO.
Therefore, we have:
0.048 moles NO * (2 moles NO2 / 2 moles NO) = 0.048 moles NO2
To convert moles of NO2 to grams, we'll multiply by the molar mass of NO2 (46.01 g/mol):
0.048 moles NO2 * (46.01 g NO2 / 1 mole NO2) = 2.208 g NO2
Therefore, 1.44 g of NO will produce 2.208 g of NO2 by the complete reaction.