Question

1. balance the chemical equation for the combustion reaction by typing the numbers in the blanks.

butane (c₄h₁₀) and oxygen (o₂):
____ c₄h₁₀ + __ o₂ → __ co₂ + ____ h₂o

1. balance the chemical equation for the combustion reaction by typing the numbers in the blanks.

benzene (c₆h₆) and oxygen (o₂):
____ c₆h₆ __ o₂ → __ co₂ + ____ h₂o

1. balance the chemical equation for the combustion reaction by typing the numbers in the blanks.

ethyne (c₂h₂) and oxygen (o₂):
____ c₂h₂ + __ o₂ → __ co₂ + ____ h₂o

Explanation:

Question 3:

Step1: Balance Carbon (C)

In \( \text{C}_4\text{H}_{10} \), there are 4 C atoms. So, we need 4 \( \text{CO}_2 \) to balance C.
Equation now: \( \underline{1} \text{C}_4\text{H}_{10} + \underline{\quad} \text{O}_2
ightarrow \underline{4} \text{CO}_2 + \underline{\quad} \text{H}_2\text{O} \)

Step2: Balance Hydrogen (H)

In \( \text{C}_4\text{H}_{10} \), there are 10 H atoms. Each \( \text{H}_2\text{O} \) has 2 H, so we need \( \frac{10}{2} = 5 \) \( \text{H}_2\text{O} \).
Equation now: \( \underline{1} \text{C}_4\text{H}_{10} + \underline{\quad} \text{O}_2
ightarrow \underline{4} \text{CO}_2 + \underline{5} \text{H}_2\text{O} \)

Step3: Balance Oxygen (O)

On the right: \( 4 \times 2 + 5 \times 1 = 8 + 5 = 13 \) O atoms. Each \( \text{O}_2 \) has 2 O, so we need \( \frac{13}{2} \), but we use whole numbers. Multiply all coefficients by 2 to eliminate fractions:
\( 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2
ightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \) (Wait, no—wait, original step1 was 1, but when we multiply by 2, let's correct. Wait, initial step1: if we keep 1 for \( \text{C}_4\text{H}_{10} \), O balance: \( 4 \times 2 + 5 \times 1 = 13 \), so \( \text{O}_2 \) is \( \frac{13}{2} \). But to make it whole, multiply all by 2:
\( 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2
ightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \). Wait, no—wait, the problem says "typing the numbers in the blanks"—maybe we can use fractions, but usually whole numbers. Wait, no, let's redo:

Wait, \( \text{C}_4\text{H}_{10} \): C=4, H=10.

Balance C: \( 4 \text{CO}_2 \).

Balance H: \( 5 \text{H}_2\text{O} \) (since 10 H / 2 = 5).

Now O: \( 4 \times 2 + 5 \times 1 = 8 + 5 = 13 \) O. So \( \text{O}_2 \) is \( \frac{13}{2} \). But to make it whole, multiply all coefficients by 2:

\( 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2
ightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \).

Wait, but maybe the problem allows fractions? No, combustion reactions use whole numbers. So the balanced equation is \( 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2
ightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \).

Wait, no—wait, initial step1: if we start with 1 \( \text{C}_4\text{H}_{10} \), then:

C: 4 → 4 \( \text{CO}_2 \).

H: 10 → 5 \( \text{H}_2\text{O} \).

O: \( 4 \times 2 + 5 \times 1 = 13 \) → \( \text{O}_2 \) is \( 13/2 \). But we can write it as \( 1 \text{C}_4\text{H}_{10} + \frac{13}{2} \text{O}_2
ightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \), but usually multiplied by 2. So the coefficients are 2, 13, 8, 10.

Question 4:

Step1: Balance Carbon (C)

In \( \text{C}_6\text{H}_6 \), there are 6 C atoms. So, we need 6 \( \text{CO}_2 \) to balance C.
Equation now: \( \underline{1} \text{C}_6\text{H}_6 + \underline{\quad} \text{O}_2
ightarrow \underline{6} \text{CO}_2 + \underline{\quad} \text{H}_2\text{O} \)

Step2: Balance Hydrogen (H)

In \( \text{C}_6\text{H}_6 \), there are 6 H atoms. Each \( \text{H}_2\text{O} \) has 2 H, so we need \( \frac{6}{2} = 3 \) \( \text{H}_2\text{O} \).
Equation now: \( \underline{1} \text{C}_6\text{H}_6 + \underline{\quad} \text{O}_2
ightarrow \underline{6} \text{CO}_2 + \underline{3} \text{H}_2\text{O} \)

Step3: Balance Oxygen (O)

On the right: \( 6 \times 2 + 3 \times 1 = 12 + 3 = 15 \) O atoms. Each \( \text{O}_2 \) has 2 O, so we need \( \frac{15}{2} \). Multiply all coefficients by 2 to eliminate fractions:
\( 2 \text{C}_6\text{H}_6 + 15 \text{O}_2
ightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O} \).

Question 5:

Step1: Balance Carbon (C)

In \( \text{C}_2\text{H}_2 \), there are 2 C atoms. So, we need 2 \( \text{CO}_2 \) to balance C.
Equation now: \( \underline{1} \text{C}_2\text{H}_2 + \underline{\quad} \text{O}_2
ightarrow \underline{2} \text{CO}_2 + \underline{\quad} \text{H}_2\text{O} \)

Step2: Balance Hydrogen (H)

In \( \text{C}_2\text{H}_2 \), there are 2 H atoms. Each \( \text{H}_2\text{O} \) has 2 H, so we need 1 \( \text{H}_2\text{O} \).
Equation now: \( \underline{1} \text{C}_2\text{H}_2 + \underline{\quad} \text{O}_2
ightarrow \underline{2} \text{CO}_2 + \underline{1} \text{H}_2\text{O} \)

Step3: Balance Oxygen (O)

On the right: \( 2 \times 2 + 1 \times 1 = 4 + 1 = 5 \) O atoms. Each \( \text{O}_2 \) has 2 O, so we need \( \frac{5}{2} \). Multiply all coefficients by 2 to eliminate fractions:
\( 2 \text{C}_2\text{H}_2 + 5 \text{O}_2
ightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \).

Answer:

s:

Question 3:

\( \boldsymbol{2} \) \( \text{C}_4\text{H}_{10} + \boldsymbol{13} \text{O}_2
ightarrow \boldsymbol{8} \text{CO}_2 + \boldsymbol{10} \text{H}_2\text{O} \)

Question 4:

\( \boldsymbol{2} \) \( \text{C}_6\text{H}_6 + \boldsymbol{15} \text{O}_2
ightarrow \boldsymbol{12} \text{CO}_2 + \boldsymbol{6} \text{H}_2\text{O} \) (Wait, no—wait, when we multiply by 2, the first coefficient is 2? Wait, no, initial step1: if we start with 1 \( \text{C}_6\text{H}_6 \), then O is \( \frac{15}{2} \), so multiply by 2: \( 2 \text{C}_6\text{H}_6 + 15 \text{O}_2
ightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O} \).

Question 5:

\( \boldsymbol{2} \) \( \text{C}_2\text{H}_2 + \boldsymbol{5} \text{O}_2
ightarrow \boldsymbol{4} \text{CO}_2 + \boldsymbol{2} \text{H}_2\text{O} \)

Wait, but maybe the problem expects the smallest whole numbers. Let's recheck:

Question 3 (Butane):

The correct balanced equation is \( 2 \text{C}_4\text{H}_{10} + 13 \text{O}_2
ightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O} \).

Question 4 (Benzene):

\( 2 \text{C}_6\text{H}_6 + 15 \text{O}_2
ightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O} \) (or simplified by dividing by 2? No, 15 is odd, so we can't. Wait, no—wait, the combustion of benzene is \( 2 \text{C}_6\text{H}_6 + 15 \text{O}_2
ightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O} \).

Question 5 (Ethyne):

\( 2 \text{C}_2\text{H}_2 + 5 \text{O}_2
ightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \) (or simplified? No, 5 is odd, so we keep it as 2, 5, 4, 2).

So the final answers are:

1. \( 2 \), \( 13 \), \( 8 \), \( 10 \)

1. \( 2 \), \( 15 \), \( 12 \), \( 6 \)

1. \( 2 \), \( 5 \), \( 4 \), \( 2 \)