Calculate the volume of oxygen required for complete burning of `90" dm"^(3)` of butane
`2C_(4)H_(10) + 13O_(2) to 8CO_(2) + 10H_(2)O`

Explanation

To calculate the volume of oxygen required for the complete burning of 90 dm³ of butane (C₄H₁₀), we will use the balanced chemical equation and apply the principles of stoichiometry.

### Step-by-step Solution:

1. **Write the Balanced Chemical Equation:**
The balanced equation for the combustion of butane is:
\[ 2C_{4}H_{10} + 13O_{2} \rightarrow 8CO_{2} + 10H_{2}O \]

2. **Identify the Volume Ratio:**
From the balanced equation, we can see that:
- 2 volumes of butane react with 13 volumes of oxygen.
This means:
\[ \text{Volume of } O_{2} = \frac{13}{2} \times \text{Volume of } C_{4}H_{10} \]

3. **Calculate the Volume of Oxygen Required:**
Given that the volume of butane is 90 dm³, we can substitute this value into the equation:
\[ \text{Volume of } O_{2} = \frac{13}{2} \times 90 \text{ dm}^3 \]

4. **Perform the Calculation:**
\[ \text{Volume of } O_{2} = \frac{13 \times 90}{2} = \frac{1170}{2} = 585 \text{ dm}^3 \]

5. **Conclusion:**
Therefore, the volume of oxygen required for the complete burning of 90 dm³ of butane is:
\[ \boxed{585 \text{ dm}^3} \]