Question

slide 5
balance a chemical equation
balance the chemical equation for the chemical reaction of burning ethanol.
inputs/reactants outputs/products
1 c₂h₅oh + 3 o₂ → 2 co₂ + 3 h₂o
carbon (c) =
oxygen (o) =
hydrogen (h) =
carbon (c) =
oxygen (o) =
hydrogen (h) =

Explanation:

Step1: Count C on reactants

In \( C_2H_5OH \), there are 2 C atoms. So reactant C count: \( 2 \).

Step2: Count C on products

In \( CO_2 \), let coefficient be \( x \). So \( x\times1 = 2 \) (from reactant C), so \( x = 2 \). Now product C: \( 2\times1 = 2 \).

Step3: Count H on reactants

In \( C_2H_5OH \), H atoms: \( 6 \) (5 from \( C_2H_5 \), 1 from \( OH \)). So reactant H count: \( 6 \).

Step4: Count H on products

In \( H_2O \), let coefficient be \( y \). \( y\times2 = 6 \), so \( y = 3 \). Now product H: \( 3\times2 = 6 \).

Step5: Count O on reactants

In \( C_2H_5OH \), O: 1; in \( O_2 \), coefficient \( z \), so O: \( 1 + 2z \).

Step6: Count O on products

In \( CO_2 \) (coefficient 2): \( 2\times2 = 4 \); in \( H_2O \) (coefficient 3): \( 3\times1 = 3 \). Total product O: \( 4 + 3 = 7 \). Wait, no, earlier steps: Wait, correct balancing: \( C_2H_5OH + 3O_2
ightarrow 2CO_2 + 3H_2O \). Now reactant O: \( 1 + 3\times2 = 7 \)? No, wait \( C_2H_5OH \) has 1 O, \( 3O_2 \) has 6 O, total 7 O. Product: \( 2CO_2 \) has 4 O, \( 3H_2O \) has 3 O, total 7 O. Correct. Now fill counts:

Reactants:

• C: 2 (from \( C_2H_5OH \))
• O: \( 1 + 3\times2 = 7 \)? Wait no, initial \( C_2H_5OH \) coefficient 1, \( O_2 \) coefficient 3. So O in reactants: 1 (from \( C_2H_5OH \)) + 3×2 (from \( O_2 \)) = 7? Wait no, let's recalculate with correct coefficients (1, 3, 2, 3):

Reactants:

• \( C_2H_5OH \): C=2, H=6, O=1
• \( 3O_2 \): O=6

Total reactants:

• C: 2
• H: 6
• O: 1 + 6 = 7

Products:

• \( 2CO_2 \): C=2, O=4
• \( 3H_2O \): H=6, O=3

Total products:

• C: 2
• H: 6
• O: 4 + 3 = 7

So filling the table:

Reactants ( \( 1 C_2H_5OH + 3 O_2 \) ):

• Carbon (C) = \( 2 \)
• Oxygen (O) = \( 1 + 3\times2 = 7 \) (or 1 (from \( C_2H_5OH \)) + 6 (from \( 3O_2 \)) = 7)
• Hydrogen (H) = \( 6 \)

Products ( \( 2 CO_2 + 3 H_2O \) ):

• Carbon (C) = \( 2\times1 = 2 \)
• Oxygen (O) = \( 2\times2 + 3\times1 = 4 + 3 = 7 \)
• Hydrogen (H) = \( 3\times2 = 6 \)

Answer:

Reactants:

• Carbon (C) = \( \boldsymbol{2} \)
• Oxygen (O) = \( \boldsymbol{7} \)
• Hydrogen (H) = \( \boldsymbol{6} \)

Products:

• Carbon (C) = \( \boldsymbol{2} \)
• Oxygen (O) = \( \boldsymbol{7} \)
• Hydrogen (H) = \( \boldsymbol{6} \)

(And the balanced equation coefficients are \( 1 C_2H_5OH + 3 O_2
ightarrow 2 CO_2 + 3 H_2O \))