Chapter 10: Problem 55
If \(1.00 \mathrm{~mol}\) of ethane gas and \(5.00 \mathrm{~mol}\) of oxygen gas react, what is the limiting reactant and how many moles of water are produced from the reaction? $$2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$
Short Answer
Expert verified
Ethane is the limiting reactant; 3.00 moles of water are produced.
Step by step solution
01
Write the Balanced Equation
The balanced chemical equation is given as \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g) + 7 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\). This provides the stoichiometric ratios of reactants and products.
02
Identify Moles Available
You are given \(1.00\) mole of ethane (\(\mathrm{C}_2 \mathrm{H}_6\)) and \(5.00\) moles of oxygen (\(\mathrm{O}_2\)). We need to determine which reactant will be exhausted first using stoichiometry.
03
Calculate Required Mole Ratio for Oxygen
According to the balanced equation, \(2\) moles of \(\mathrm{C}_2 \mathrm{H}_6\) react with \(7\) moles of \(\mathrm{O}_2\). Therefore, \(1.00\) mole of \(\mathrm{C}_2 \mathrm{H}_6\) requires \(\frac{7}{2} = 3.50\) moles of \(\mathrm{O}_2\).
04
Determine Limiting Reactant
You have \(5.00\) moles of \(\mathrm{O}_2\) available, more than the \(3.50\) moles required for burning \(1.00\) mole of \(\mathrm{C}_2 \mathrm{H}_6\). Since you have enough \(\mathrm{O}_2\), \(\mathrm{C}_2 \mathrm{H}_6\) is the limiting reactant.
05
Calculate Moles of Water Produced
From the balanced equation, \(2\) moles of \(\mathrm{C}_2 \mathrm{H}_6\) produce \(6\) moles of \(\mathrm{H}_2 \mathrm{O}\). Therefore, \(1.00\) mole of \(\mathrm{C}_2 \mathrm{H}_6\) produces \(\frac{6}{2} = 3.00\) moles of \(\mathrm{H}_2 \mathrm{O}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry is like the recipe of a chemical reaction. It tells you exactly how much of each ingredient, or reactant, you need to create a certain amount of the finished product, the product. This concept is crucial because it helps ensure that nothing is wasted or missing when a reaction occurs. In a chemical reaction, stoichiometry involves calculating the proportions of elements and compounds involved.
Understanding stoichiometry allows chemists to predict quantities of products formed in reactions. For example, if you know how much of a reactant you start with, stoichiometry helps predict how much of your product you will make. This requires using a balanced chemical equation, which we'll discuss next.
Understanding stoichiometry allows chemists to predict quantities of products formed in reactions. For example, if you know how much of a reactant you start with, stoichiometry helps predict how much of your product you will make. This requires using a balanced chemical equation, which we'll discuss next.
Balanced Chemical Equation
A balanced chemical equation is like a set of rules for the reaction. It shows which reactants take part and what products are formed. Most importantly, it respects the law of conservation of mass, meaning matter isn't created or destroyed during the reaction.
For example, in the reaction:\[2 \mathrm{C}_{2} \mathrm{H}_{6}(g) + 7 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\]the equation shows that 2 molecules of ethane react with 7 molecules of oxygen to produce 4 molecules of carbon dioxide and 6 molecules of water. Each side of the equation has the same number of each type of atom. It is balanced!
For example, in the reaction:\[2 \mathrm{C}_{2} \mathrm{H}_{6}(g) + 7 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\]the equation shows that 2 molecules of ethane react with 7 molecules of oxygen to produce 4 molecules of carbon dioxide and 6 molecules of water. Each side of the equation has the same number of each type of atom. It is balanced!
- Balance the number of atoms of each element on both sides of the equation.
- Ensure that each type of atom appears in equal numbers, confirming the equation adheres to conservation of mass.
Mole Ratio
The mole ratio comes from the coefficients in the balanced chemical equation. It tells us the proportions of reactants and products that take part in the reaction. It’s an essential aspect of stoichiometry to solve limiting reactant problems.
Using our example, if we take the balanced equation mentioned above, the mole ratio tells us that:
Using our example, if we take the balanced equation mentioned above, the mole ratio tells us that:
- 2 moles of ethane (\(\mathrm{C}_2 \mathrm{H}_6\)) react with 7 moles of oxygen (\(\mathrm{O}_2\)).
- Produces 6 moles of water (\(\mathrm{H}_2 \mathrm{O}\)).
Chemical Reaction
A chemical reaction involves the transformation of substances when reactants convert into products. It is a process that involves breaking and forming chemical bonds, leading to a change in the chemical composition of the substances involved.
In our example reaction, ethane and oxygen combine to form carbon dioxide and water. These processes are distinct as:
In our example reaction, ethane and oxygen combine to form carbon dioxide and water. These processes are distinct as:
- Reactants (substances you start with) are on the left side of the equation.
- Products (substances produced) are on the right side of the equation.