To balance the given chemical equations and classify them, let's work through each equation step-by-step.
d) C2H6 + O2 → CO2 + H2O
Balancing the equation:
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Count the number of each type of atom on both sides:
- Reactants: C (2), H (6), O (2)
- Products: C (1 for CO2), H (2 for H2O), O (3 total: 2 for CO2 + 1 for H2O)
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Balance carbon (C):
- Place a coefficient of 2 in front of CO2: \[ C2H6 + O2 → 2CO2 + H2O \]
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Balance hydrogen (H):
- Place a coefficient of 3 in front of H2O: \[ C2H6 + O2 → 2CO2 + 3H2O \]
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Balance oxygen (O):
- Now we have 4 O (from 2 CO2) + 3 O (from 3 H2O) = 7 O total on the products side.
- To get 7 O from O2 (which gives 2 O per molecule), we need \( \frac{7}{2} \) or 3.5 O2. To eliminate the fraction, we can multiply the entire equation by 2: \[ 2C2H6 + 7O2 → 4CO2 + 6H2O \]
Final balanced equation: \[ 2C2H6 + 7O2 → 4CO2 + 6H2O \]
Classification:
- This reaction is a combustion reaction since it involves the burning of an alkane (C2H6) in the presence of oxygen.
e) NO + O2 → NO2
Balancing the equation:
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Count the number of each type of atom on both sides:
- Reactants: N (1 for NO), O (2)
- Products: N (1 for NO2), O (2 total: 1 from NO2 for 1 N and 1 from O2)
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Since nitrogen (N) and oxygen (O) are already balanced, we can write: \[ NO + O2 → NO2 \]
Final balanced equation: \[ 2NO + O2 → 2NO2 \]
Classification:
- This reaction is an oxidation-reduction (redox) reaction as nitrogen monoxide (NO) is oxidized to nitrogen dioxide (NO2) in the presence of oxygen.
Summary:
- d) 2C2H6 + 7O2 → 4CO2 + 6H2O - Combustion reaction
- e) 2NO + O2 → 2NO2 - Oxidation-reduction (redox) reaction