To determine the number of moles of MgO produced when 0.20 mole of O2 reacts completely, we need to first write and balance the chemical reaction. Once we have the balanced equation, we can use the stoichiometric mole ratios to find the moles of MgO produced.
Writing and Balancing the Chemical Reaction
The reaction between magnesium and oxygen produces magnesium oxide (MgO). The unbalanced chemical equation is:
Mg + O2 → MgO
To balance this equation, we need to determine the coefficients needed before each compound to satisfy the law of conservation of mass. There are 2 oxygens on the reactant side but only 1 oxygen on the product side. To fix this, we place a 2 before the MgO:
Mg + O2 → 2MgO
Now there are 2 magnesiums on the product side but only 1 on the reactant side. To balance this out, we place a 2 before the Mg:
2Mg + O2 → 2MgO
The balanced chemical equation is:
2Mg + O2 → 2MgO
Using Stoichiometry to Find Moles of MgO
Now that we have the balanced equation, we can use stoichiometry to determine the moles of MgO produced from 0.20 mole of O2. Stoichiometry relates the quantities in a balanced chemical equation.
First, we need to look at the mole ratios from the coefficients in the balanced equation:
- 2 moles of Mg reacts with 1 mole of O2
- 2 moles of MgO is produced for every 1 mole of O2
We are given 0.20 mole of O2 initially. Using the mole ratio between O2 and MgO, if 0.20 mole of O2 reacts, it will produce 2 times as many moles of MgO:
0.20 mole O2 x (2 mole MgO/1 mole O2) = 0.40 mole MgO
Therefore, when 0.20 mole of O2 reacts completely, 0.40 mole of MgO is produced.
Conclusion
By writing and balancing the chemical equation for the reaction of magnesium with oxygen, and then using stoichiometric mole ratios, we determined that 0.20 mole of O2 reacting completely will produce 0.40 mole of MgO.
The key steps were:
- Write and balance the chemical equation: 2Mg + O2 → 2MgO
- Use the mole ratio between O2 and MgO based on the balanced equation coefficients
- Apply the mole ratio to the given 0.20 mole O2 to calculate the moles of MgO produced
Using stoichiometric calculations allows us to quantify the amount of products formed in a chemical reaction when we know the amount of one reactant.
Practice Examples
Let’s try some more practice problems using stoichiometry and mole ratios:
Problem 1
Calculate the moles of NH3 produced when 0.50 mole of N2 reacts with excess H2 according to the equation:
N2 + 3H2 → 2NH3
Solution:
- The balanced equation shows 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3
- We are given 0.50 mole of N2 initially
- Using the mole ratio between N2 and NH3:
- 0.50 mole N2 x (2 mole NH3/1 mole N2) = 1.0 mole NH3
- Therefore, 0.50 mole of N2 produces 1.0 mole of NH3 when reacting with excess H2
Problem 2
What mass of CO2 is formed when 40.0 g CaCO3 decomposes according to the chemical equation:
CaCO3 → CaO + CO2
Solution:
- First, convert 40.0 g CaCO3 to moles using molar mass (MM CaCO3 = 100.1 g/mol):
- 40.0 g CaCO3 x (1 mole CaCO3/100.1 g CaCO3) = 0.399 mole CaCO3
- The balanced equation shows 1 mole of CaCO3 produces 1 mole of CO2
- Use the mole ratio:
- 0.399 mole CaCO3 x (1 mole CO2/1 mole CaCO3) = 0.399 mole CO2
- Convert moles of CO2 to grams using molar mass (MM CO2 = 44.01 g/mol):
- 0.399 mole CO2 x (44.01 g CO2/1 mole CO2) = 17.6 g CO2
Therefore, 40.0 g CaCO3 decomposing produces 17.6 g CO2
Stoichiometry Concepts
Let’s go over some of the important concepts related to stoichiometric calculations:
Mole Ratios
Mole ratios relating amounts of reactants and products come from the coefficients in a balanced chemical equation. They allow conversion between quantities using ratios.
Limiting Reactant
The limiting reactant is the reactant that gets used up first in a reaction and limits the amounts of products formed. The mole ratio calculates the maximum amount of product possible from the limiting reactant.
Excess Reactant
The reactant present in excess does not limit product formation. The excess amount remains unreacted after the limiting reactant is consumed.
Percent Yield
The percent yield is the actual yield of a reaction divided by the theoretical yield from stoichiometry. It represents the efficiency of the chemical reaction.
Percent Yield = (Actual Yield/Theoretical Yield) x 100%
Stoichiometry Problem Solving
Here are some step-by-step strategies for using stoichiometry to solve various chemistry problems:
Finding a Product Amount
- Write a balanced chemical equation for the reaction
- Convert from grams to moles of the known reactant amount using molar mass
- Determine the mole ratio between the known reactant and unknown product from the balanced equation
- Multiply the moles of known reactant by the mole ratio to find moles of unknown product
- If needed, convert moles of product to grams using its molar mass
Finding a Reactant Amount
- Write a balanced chemical equation
- Convert from grams to moles of known product amount
- Determine mole ratio between known product and unknown reactant
- Multiply moles of known product by the inverted mole ratio to find moles of unknown reactant
- If needed, convert moles of reactant to grams using its molar mass
Determining a Limiting Reactant
- Convert all reactant amounts to moles
- Divide each reactant moles by its coefficient in the balanced equation to find mole ratio
- The reactant with the lowest mole ratio is the limiting reactant
Calculating Percent Yield
- Determine theoretical yield from stoichiometric calculations
- Determine actual yield from experimental results
- Divide actual yield by theoretical yield and multiply by 100% to find percent yield
Practice using these strategies to become proficient at stoichiometry problems!
Real-World Applications
Here are some examples of using stoichiometry calculations in real-world chemistry:
Industrial Production
Chemical engineers use stoichiometry to determine optimal reactant ratios and predict the maximum yields for large-scale industrial production.
Greenhouse Gas Emissions
Scientists can calculate the amount of carbon dioxide produced from the combustion of fossil fuels based on chemical reactions and fuel consumption data.
Limiting Nutrients in Agriculture
Farmers apply stoichiometric principles to identify limiting plant nutrients in soil and determine optimal fertilizer amounts.
Medicine and Pharmacology
Medical professionals use stoichiometry to calculate drug and medication dosages based on proportional reactions in the human body.
Food Science and Nutrition
Nutrition facts labels are based on using stoichiometry to quantitatively analyze biochemical reactions and energy content of foods.
In all these examples, stoichiometry provides a powerful quantitative technique for analyzing chemical processes and making data-driven decisions.
Stoichiometry Problem Set
Practice your stoichiometry skills with this set of math problems:
Problem 1
What is the theoretical yield of ammonia (NH3) in grams if 10.0 g of nitrogen (N2) is reacted with excess hydrogen (H2) according to the following balanced equation?
N2 + 3H2 → 2NH3
Problem 2
How many grams of chlorine gas (Cl2) are produced when 20.0 g of MnO2 reacts with excess hydrochloric acid (HCl) according to the balanced equation below?
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
Problem 3
What is the percent yield if an experiment produced 2.50 g of sodium sulfate product (Na2SO4) when 10.0 g of sodium chloride (NaCl) reacted with sulfuric acid according to the following unbalanced reaction?
NaCl + H2SO4 → Na2SO4 + HCl
Problem 4
For the reaction:
4NH3 + 7O2 → 4NO2 + 6H2O
If 25.0 g of NH3 is reacted with excess O2, what mass of NO2 is produced theoretically?
Problem 5
If 5.00 g of Mg is reacted with 2.00 g of O2 in an experiment, determine which reactant is limiting and calculate the theoretical yield of MgO.
Mg + O2 → MgO
Solutions to Stoichiometry Problems
Problem 1
- Balanced equation: N2 + 3H2 → 2NH3
- 10.0 g N2 x (1 mole N2/28.0 g N2) = 0.357 mole N2
- Mole ratio from balanced equation:
- 0.357 mole N2 x (2 mole NH3/1 mole N2) = 0.714 mole NH3
- 0.714 mole NH3 x (17.0 g NH3/1 mole NH3) = 12.2 g NH3
The theoretical yield is 12.2 g NH3
Problem 2
- Balanced equation: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
- 20.0 g MnO2 x (1 mole MnO2/86.9 g MnO2) = 0.230 mole MnO2
- Mole ratio from balanced equation:
- 0.230 mole MnO2 x (1 mole Cl2/1 mole MnO2) = 0.230 mole Cl2
- 0.230 mole Cl2 x (70.9 g Cl2/1 mole Cl2) = 16.3 g Cl2
The theoretical yield is 16.3 g Cl2
Problem 3
- Balanced equation: 2NaCl + H2SO4 → Na2SO4 + 2HCl
- Theoretical yield:
- 10.0 g NaCl x (1 mole NaCl/58.5 g NaCl) = 0.171 mole NaCl
- Mole ratio from balanced equation:
- 0.171 mole NaCl x (1 mole Na2SO4/2 mole NaCl) = 0.0855 mole Na2SO4
- 0.0855 mole Na2SO4 x (142.0 g Na2SO4/1 mole Na2SO4) = 12.1 g Na2SO4
- Actual yield = 2.50 g Na2SO4
- Percent Yield = (Actual/Theoretical) x 100% = (2.50 g/12.1 g) x 100% = 20.7%
The percent yield is 20.7%
Problem 4
- Balanced equation: 4NH3 + 7O2 → 4NO2 + 6H2O
- 25.0 g NH3 x (1 mole NH3/17.0 g NH3) = 1.47 mole NH3
- Mole ratio from balanced equation:
- 1.47 mole NH3 x (4 mole NO2/4 mole NH3) = 1.47 mole NO2
- 1.47 mole NO2 x (46.0 g NO2/1 mole NO2) = 67.6 g NO2
The theoretical yield is 67.6 g NO2
Problem 5
- Balanced equation: Mg + O2 → MgO
- 5.00 g Mg x (1 mole Mg/24.3 g Mg) = 0.206 mole Mg
- 2.00 g O2 x (1 mole O2/32.0 g O2) = 0.0625 mole O2
- Mole ratio:
- Mg: 0.206 mole Mg/1 mole Mg per equation = 0.206
- O2: 0.0625 mole O2/1 mole O2 per equation = 0.0625
- O2 is the limiting reactant since it has the lower mole ratio
- Using O2:
- 0.0625 mole O2 x (1 mole MgO/1 mole O2) = 0.0625 mole MgO
- 0.0625 mole MgO x (40.3 g MgO/mole MgO) = 2.52 g MgO
O2 is limiting. The theoretical yield is 2.52 g MgO.
Conclusion
In summary, stoichiometry and mole ratios allow chemists to quantitatively relate reactant amounts to product amounts using balanced chemical equations. It is an essential tool for determining yields, identifying limiting reactants, calculating percentage yield, and quantitative chemical analysis. Practice thinking stoichiometrically and using mole ratios to answer chemistry questions.
