Question

1. balance the equation for the combustion of ethanol by dragging the coefficients into the equation.

? c₂₅h₅₂ + ? o₂ → ? co₂ + ? h₂o

Explanation:

Step1: Balance Carbon atoms

In \( \ce{C_{25}H_{52}} \), there are 25 C atoms. So, \( \ce{CO_2} \) should have a coefficient of 25 to balance C.
\( \ce{C_{25}H_{52}} + \ce{O_2}
ightarrow 25\ce{CO_2} + \ce{H_2O} \)

Step2: Balance Hydrogen atoms

In \( \ce{C_{25}H_{52}} \), there are 52 H atoms. Each \( \ce{H_2O} \) has 2 H atoms, so the coefficient for \( \ce{H_2O} \) is \( \frac{52}{2} = 26 \).
\( \ce{C_{25}H_{52}} + \ce{O_2}
ightarrow 25\ce{CO_2} + 26\ce{H_2O} \)

Step3: Balance Oxygen atoms

On the right side, \( 25\ce{CO_2} \) has \( 25 \times 2 = 50 \) O atoms, and \( 26\ce{H_2O} \) has \( 26 \times 1 = 26 \) O atoms. Total O on right: \( 50 + 26 = 76 \). Each \( \ce{O_2} \) has 2 O atoms, so the coefficient for \( \ce{O_2} \) is \( \frac{76}{2} = 38 \).
\( \ce{C_{25}H_{52}} + 38\ce{O_2}
ightarrow 25\ce{CO_2} + 26\ce{H_2O} \)

Answer:

The balanced equation is \( \boldsymbol{1}\ce{C_{25}H_{52}} + \boldsymbol{38}\ce{O_2}
ightarrow \boldsymbol{25}\ce{CO_2} + \boldsymbol{26}\ce{H_2O} \) (coefficients: 1, 38, 25, 26)