1. Balanced reaction

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

So,

• $1$ mol butane gives $5$ mol water
• $1$ mol butane needs $\dfrac{13}{2} = 6.5$ mol $\mathrm{O}_2$

2. Molar masses

For butane, $\mathrm{C}_4\mathrm{H}_{10}$:

$M = 4(12) + 10(1) = 48 + 10 = 58\ \mathrm{g\,mol^{-1}}$

For oxygen, $\mathrm{O}_2$:

$M = 2(16) = 32\ \mathrm{g\,mol^{-1}}$

3. Convert given masses to moles

Butane

$174.0\ \mathrm{kg} = 174000\ \mathrm{g}$

$n(\mathrm{C}_4\mathrm{H}_{10}) = \frac{174000}{58} = 3000\ \mathrm{mol}$

Oxygen

$320.0\ \mathrm{kg} = 320000\ \mathrm{g}$

$n(\mathrm{O}_2) = \frac{320000}{32} = 10000\ \mathrm{mol}$

4. Find the limiting reagent

Required oxygen for $3000$ mol butane:

$3000 \times 6.5 = 19500\ \mathrm{mol\ O_2}$

Available oxygen = $10000$ mol only.

Hence, $\mathrm{O}_2$ is the limiting reagent.

5. Moles of butane that can react with available oxygen

From the equation,

$6.5\ \text{mol } \mathrm{O}_2 \text{ reacts with } 1\ \text{mol butane}$

So butane consumed:

$n(\mathrm{C}_4\mathrm{H}_{10}\ \text{reacted}) = \frac{10000}{6.5} = \frac{20000}{13}\ \mathrm{mol}$

6. Moles of water formed

From stoichiometry,

$1\ \text{mol butane} \rightarrow 5\ \text{mol water}$

Thus,

$n(\mathrm{H_2O}) = 5 \times \frac{20000}{13} = \frac{100000}{13} \approx 7692.31\ \mathrm{mol}$

7. Mass of water formed

Molar mass of water:

$M(\mathrm{H_2O}) = 2(1) + 16 = 18\ \mathrm{g\,mol^{-1}}$

So,

$m(\mathrm{H_2O}) = \frac{100000}{13} \times 18 = \frac{1800000}{13}\ \mathrm{g} \approx 138461.54\ \mathrm{g}$

8. Convert mass of water to volume

Given density of water = $1\ \mathrm{g\,mL^{-1}}$.

Therefore,

$138461.54\ \mathrm{g} = 138461.54\ \mathrm{mL} = 138.46154\ \mathrm{L}$

Nearest integer:

$\boxed{138}$

9. Comparison with stored answer

Stored correct answer = $138$.

Our derived answer also is $138$, so they agree.