The chemical reaction between butane ($\mathrm{C}_4\mathrm{H}_{10}$) and oxygen ($\mathrm{O}_2$) to produce carbon dioxide ($\mathrm{CO}_2$) and water ($\mathrm{H}_2\mathrm{O}$) can be represented by the following equation:

$ \mathrm{C}_4\mathrm{H}_{10}(\mathrm{g}) + \frac{13}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{g}) + 5 \mathrm{H}_2\mathrm{O}(\mathrm{l}) $

Given that we have 174.0 kg of butane and 320.0 kg of oxygen, we are tasked with calculating the volume of water produced in liters.

Mole Calculation for Water:

The stoichiometry of the reaction shows that 5 moles of water are produced per 13/2 moles of oxygen.

Using the moles of oxygen available from input, calculate the moles of water produced:

$ \text{Moles of } \mathrm{H}_2\mathrm{O} = 5 \times \frac{2}{13} \times (10 \times 10^3) $

Mass of Water Produced:

Calculate the mass of water using the molar mass of water (18 g/mol):

$ \text{Mass of } \mathrm{H}_2\mathrm{O} = \left(\frac{10^5}{13}\right) \times 18 $

The result will be:

$ = 1.3846 \times 10^5 \ \text{g} $

Volume of Water:

Convert the mass of water to volume, considering the density of water is $1 \ \text{g/mL}$ (or $1 \ \text{g/L}$):

$ \text{Volume of } \mathrm{H}_2\mathrm{O} = 138.46 \ \text{liters} $

Therefore, the volume of water formed is approximately 138 liters when rounded to the nearest integer.