Question
Ch4 + 2O2 → CO2 + 2H2O
If you change 2o2 to a 302, what will be the change in the results on the right side of the equation.
A. There are now 3CO2 molecules
B. There is an O2 molecule left over
C. Nothing changes in the equation
D. There are now 3H2O molecules
If you change 2o2 to a 302, what will be the change in the results on the right side of the equation.
A. There are now 3CO2 molecules
B. There is an O2 molecule left over
C. Nothing changes in the equation
D. There are now 3H2O molecules
Answers
GPT-4o mini
To analyze the given reaction:
Initially:
\[ \text{C}_4\text{H}_{10} + 2 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \]
If you change \( 2 \text{O}_2 \) to \( 3 \text{O}_2 \), the equation becomes:
\[ \text{C}_4\text{H}_{10} + 3 \text{O}_2 \rightarrow ? \]
Now, let's balance the reaction with the new amount of \( \text{O}_2 \):
1. **Using \( 3 \text{O}_2 \)**:
When we react \( 3 \text{O}_2 \) with the alkane, the reaction will produce more \( \text{CO}_2 \) than before (when only \( 2 \text{O}_2 \) were used).
To find out how many \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) molecules would be produced, we can observe the stoichiometry:
- 1 \( \text{C}_4\text{H}_{10} \) molecule reacts with \( 3 \) \( \text{O}_2 \).
- Combustion of \( \text{C}_4\text{H}_{10} \) yields \( 4 \) \( \text{CO}_2 \) and a certain number of \( \text{H}_2\text{O} \) molecules.
By changing from \( 2O_2 \) to \( 3O_2 \):
- We would still fully combust \( \text{C}_4\text{H}_{10} \) if sufficient \( \text{O}_2 \) is present and produce the same amount of \( \text{CO}_2 \).
- However, you have an excess of \( \text{O}_2 \), which means there is one \( O_2 \) molecule left unreacted after the reaction.
Thus, the correct answer is:
**B. There is an O2 molecule left over.**
Initially:
\[ \text{C}_4\text{H}_{10} + 2 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \]
If you change \( 2 \text{O}_2 \) to \( 3 \text{O}_2 \), the equation becomes:
\[ \text{C}_4\text{H}_{10} + 3 \text{O}_2 \rightarrow ? \]
Now, let's balance the reaction with the new amount of \( \text{O}_2 \):
1. **Using \( 3 \text{O}_2 \)**:
When we react \( 3 \text{O}_2 \) with the alkane, the reaction will produce more \( \text{CO}_2 \) than before (when only \( 2 \text{O}_2 \) were used).
To find out how many \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) molecules would be produced, we can observe the stoichiometry:
- 1 \( \text{C}_4\text{H}_{10} \) molecule reacts with \( 3 \) \( \text{O}_2 \).
- Combustion of \( \text{C}_4\text{H}_{10} \) yields \( 4 \) \( \text{CO}_2 \) and a certain number of \( \text{H}_2\text{O} \) molecules.
By changing from \( 2O_2 \) to \( 3O_2 \):
- We would still fully combust \( \text{C}_4\text{H}_{10} \) if sufficient \( \text{O}_2 \) is present and produce the same amount of \( \text{CO}_2 \).
- However, you have an excess of \( \text{O}_2 \), which means there is one \( O_2 \) molecule left unreacted after the reaction.
Thus, the correct answer is:
**B. There is an O2 molecule left over.**