Stoichiometry help #4

Question

Nitrogen combines with oxygen in the atmosphere during lightning flashes to form nitrogen monoxide, NO, which then reacts further with O₂ to produce nitrogen dioxide, NO₂.

(a) What mass of NO₂ is formed when NO reacts with 364 g O₂?

__ g

(b) How many grams of NO are required to react with this amount of O₂?

__ g

Answer 1

First, we must write out the two equations:

Nitrogen and oxygen are both diatomic elements, meaning their chemical formulas consist of two of the same molecules bonded together -- N₂ and O₂. The formulas for nitrogen oxide and nitrogen dioxide are provided in the question -- NO and NO₂.

Now that we know how to write the chemical formulas for all elements involved, we write the unbalanced equations:

1. N₂ + O₂ --> NO (unbalanced)
2. NO + O₂ --> NO₂ (unbalanced)

1. To balance the first equation, we can simply put a coefficient of 2 in front of the product NO:

N₂ + O₂ --> 2NO (balanced)

2. To balance the second equation, we need a little more work. There are 3 O atoms on the left and 2 O atoms on the right. We can make our O atoms on the left an even number by placing a coefficient of 2 in front of the NO molecule:

2NO + O₂ --> NO₂ (unbalanced)

Now, we must balance the N atoms by placing a coefficient of 2 in front of the product NO₂:

2NO + O₂ --> 2NO₂ (balanced)

Now, we have our 2 balanced equations:

1. N₂ + O₂ --> 2NO
2. 2NO + O₂ --> 2NO₂

(a) The question then asks how much NO₂ is formed when NO reacts with a certain mass (364 g) of O₂. This matches our second equation above, and thus we can use that equation to help us solve.

In stoichiometry, it is best to convert mass to moles first, use the equation for the mole-to-mole conversions, and then convert moles back to mass (as needed). We start with our given amount, and work our way towards the solution:

364 g O₂ X 1 mol O_{2 =} 11.375 mol O₂ X 2 mol NO_{2 =} 22.75 mol NO₂ formed X 46 g NO_{2 =}

32 g O₂ 1 mol O₂ 1 mol NO₂

1,046.5 g NO₂ (product) formed

(b) The second part of the question asks how many grams of NO are required for this amount of O_2. Again, we start with our given amount of 364 g O₂, except now we are solving for grams of NO rather than grams of NO_2. The process is the same as above in (a):

364 g O₂ X 1 mol O_{2 =} 11.375 mol O₂ X 2 mol NO ₌ 22.75 mol NO required X 30 g NO ₌

32 g O₂ 1 mol O₂ 1 mol NO

682.5 g NO (reactant) required