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Given the equation below, what substance will be produced at a faster rate?

[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]

A. \(\text{CO}_2\)

B. \(\text{CO}_2\) and \(\text{H}_2\text{O}\) at equal rates

C. \(\text{H}_2\text{O}\)

D. [tex]\(\text{C}_3\text{H}_8\)[/tex]



Answer :

GinnyAnswer
To determine which substance will be produced at a faster rate in the given chemical reaction:

[tex]\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]

we need to look at the coefficients of the products (CO₂ and H₂O) in the balanced chemical equation.

### Step-by-Step Analysis:

1. Identify the Reactants and Products:
- Reactants: \(C_3H_8\) and \(O_2\)
- Products: \(CO_2\) and \(H_2O\)

2. Write Down the Balanced Equation:
[tex]\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]

3. Observe the Coefficients of the Products:
- For \(CO_2\), the coefficient is 3.
- For \(H_2O\), the coefficient is 4.

4. Compare the Coefficients:
- The coefficient for \(H_2O\) (4) is greater than the coefficient for \(CO_2\) (3).

### Conclusion:
Since the coefficient of \(H_2O\) is larger than that of \(CO_2\), \(H_2O\) will be produced at a faster rate than \(CO_2\).

So, the substance that will be produced at a faster rate is:

[tex]\[ \boxed{H_2O} \][/tex]

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