Stoichiometry: finding the yield in reaction of sodium azide with iron oxide

In summary: However, when I do the calculation with the exact masses, the decimal goes into the wrong place and I get a higher yield. Anyway, the yield is 89.2%.
  • #1
pc2-brazil
205
3
Thank you in advance.

Homework Statement



Nitrogen gas can be obtained from the reaction between sodium azide and iron oxide.

[tex]6NaN_3_{(s)}+Fe_2O_3_{(s)} \rightarrow 3Na_2O_{(s)}+2Fe_{(s)}+9N_2_{(g)}[/tex]

a) What is the yield of this reaction, knowing that, from 390 g of azide and 400 g of iron oxide, 100 g of metalic iron were produced?

b) What is the volume (in litres = L) of N2 produced, in a pression of 0.82 atm and a temperature of 27oC, in the conditions given in item a?

Universal constant of gases = R = 0.082 (atm L)/(mol K).

Homework Equations



[tex]PV = nRT[/tex] (P = atm, V = litres, n = mol, R = universal constant of gases, T = Kelvin).

The Attempt at a Solution



a) First, the molar relation between sodium azide and iron oxide, in order to discover what is the limiting reactant:

Mass of 6NaN3 = 6(23 + 3(16)) = 6(23 + 48) = 6(65) = 390 g.
Mass of Fe2O3 = 2(56) + 3(16) = 112 + 48 = 160 g.
Mass of 2Fe = 2(56) = 112 g.
[tex]6 mol NaN_3 \rightarrow 1 mol Fe_2O_3[/tex]
[tex]390 g NaN_3\rightarrow 160 g Fe_2O_3[/tex]

Since we already have 390 g of NaN3, a rule of three will not be necessary.
This mass of sodium azide produces 160 g of iron oxide; thus, there is excess of iron oxide. Therefore, the limiting reactant is NaN3.

Mass of metalic iron produced:
[tex]6 mol NaN_3 \rightarrow 2 mol Fe[/tex]
[tex]390 g \rightarrow 112 g[/tex]
The mass of metalic iron is 112g. Since the problem states the mass produced is 100 g, dividing 100 / 112 will give the yield (R):
[tex]R = \frac{100}{112}[/tex]
R equals approximately 89.2%.

b) Convert 27oC to Kelvin = 27 + 273 = 300 K.
To find the molar volume (Vm): [tex]PV_m = nRT[/tex], then [tex]0.82V_m = 0.082 \times 300[/tex]; thus Vm = 30 L.
Since 390 g of NaN3 equal 6 mol, then the volume of N2 produced is 9Vm.
[tex]V = 9V_m = 9 \times 30 = 270 L[/tex]
But the yield is 89.2%:
[tex]V = 0.892 \times 270[/tex]
V equals 240.82 L.
 
Physics news on Phys.org
  • #2
a OK, for b I got a little bit more. Not much. Don't round down intermediate results.
 
  • #3
Borek said:
a OK, for b I got a little bit more. Not much. Don't round down intermediate results.

Thank you for the response. This calculation was done by hand (100/112 = 50/58 = 25/28 = ~89,2). The next digit would be 8, which makes difference in the result (270 * 89.28 = 271.056). Anyway, if we were to do the calculation of the volume like this:
[tex]V = \frac{100 \times 270}{112} = \frac{25 * 270}{28} = \frac{25 * 135}{14}[/tex]
the result would be better.
It is very good to know that the reasoning to solve the problem is right.
 
  • #4
It is possible that part of the difference is in molar masses, I am using my stoichiometry calculator and it does all calculations using as exact molar masses as possible.
 

1. How do you calculate the yield in a reaction using stoichiometry?

In order to calculate the yield in a reaction using stoichiometry, you need to know the balanced chemical equation for the reaction. From there, you can use the molar ratios of the reactants and products to determine the theoretical yield, which is the maximum amount of product that can be formed. Then, you can compare the actual yield from the reaction to the theoretical yield to calculate the percent yield.

2. What is the purpose of using stoichiometry in chemistry?

Stoichiometry is used in chemistry to determine the quantitative relationships between reactants and products in a chemical reaction. It allows for the calculation of the amount of product that can be formed from a given amount of reactants, as well as the amount of reactants needed to produce a desired amount of product. This information is crucial in industries such as manufacturing and pharmaceuticals, where precise control over the amount of product is necessary.

3. Can stoichiometry be used to predict the outcome of a chemical reaction?

Stoichiometry can be used to predict the outcome of a chemical reaction to some extent. By determining the molar ratios of the reactants and products, it can give an idea of the relative amounts of each substance that will be involved in the reaction. However, there are many other factors that can influence the outcome of a reaction, such as temperature, pressure, and catalysts, which may not be accounted for in stoichiometric calculations.

4. Why is it important to use a balanced chemical equation in stoichiometry?

A balanced chemical equation is necessary for accurate stoichiometric calculations because it represents the correct ratios of reactants and products involved in the reaction. Without a balanced equation, the molar ratios would be incorrect, leading to inaccurate calculations of the yield and percent yield.

5. What are the limitations of using stoichiometry to find the yield in a chemical reaction?

There are several limitations to using stoichiometry to find the yield in a chemical reaction. One limitation is that it assumes all reactants are completely consumed and converted to products, which may not always be the case in a real-world scenario. Additionally, stoichiometry does not account for side reactions or impurities that may affect the yield. It also assumes that the reaction takes place under ideal conditions, which may not always be the case in a laboratory setting.

Similar threads

  • Biology and Chemistry Homework Help
Replies
4
Views
2K
Replies
3
Views
14K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
4K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
8K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
9K
  • Biology and Chemistry Homework Help
Replies
2
Views
4K
Back
Top