Question

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

?c₂₅h₅₂ + ?o₂ → ?co₂ + ?h₂o
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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.

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 \( \frac{52}{2} = 26 \).

Step3: Balance Oxygen atoms

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

Answer:

1 \( \ce{C_{25}H_{52}} + 38 \ce{O_2}
ightarrow 25 \ce{CO_2} + 26 \ce{H_2O} \)