Butane reacts with oxygen to produce carbon dioxide and water following the equation given below
$$C_4H_{10}(g) + \frac{13}{2}O_2(g) \to 4CO_2(g) + 5H_2O(l)$$.
If 174.0 kg of butane is mixed with 320.0 kg of $$O_2$$, the volume of water formed in litres is _____.(Nearest integer)
[Given : (a) Molar mass of C, H, O are 12, 1, 16 g $$mol^{-1}$$ respectively, (b) Density of water = 1 g $$mL^{-1}$$] 

Molar masses (in $$\mathrm{g\;mol^{-1}}$$): $$C_4H_{10}=58$$, $$O_2 = 32$$, $$H_2O = 18$$.

Moles of butane mixed
$$n_{C_4H_{10}}=\frac{174.0\times10^{3}}{58}=3000\;\text{mol}$$

Moles of oxygen mixed
$$n_{O_2}=\frac{320.0\times10^{3}}{32}=10000\;\text{mol}$$

Balanced equation (fractional coefficients as given) :
$$C_4H_{10}+ \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$$

Stoichiometric ratios
$$1\;\text{mol}\;C_4H_{10} \;:\; \frac{13}{2}=6.5\;\text{mol}\;O_2$$
$$1\;\text{mol}\;C_4H_{10} \;:\; 5\;\text{mol}\;H_2O$$

Oxygen needed for the available butane
$$3000 \times 6.5 = 19500\;\text{mol}\;O_2$$

Only $$10000\;\text{mol}\;O_2$$ are present, so $$O_2$$ is the limiting reagent.

From the equation, $$6.5\;\text{mol}\;O_2 \rightarrow 5\;\text{mol}\;H_2O$$, or
$$n_{H_2O} = n_{O_2}\times\frac{5}{6.5} = 10000\times\frac{5}{6.5} = 10000\times\frac{10}{13} = 7692.3077\;\text{mol}$$

Mass of water formed
$$m_{H_2O} = 7692.3077 \times 18 = 1.3846 \times 10^{5}\;\text{g} = 138.46\;\text{kg}$$

Density of liquid water $$\approx 1\;\text{kg L}^{-1}$$, therefore
$$V_{H_2O} = 138.46\;\text{kg}\times\frac{1\;\text{L}}{1\;\text{kg}} \approx 138\;\text{L}$$

Hence, the volume of water produced is about $$\mathbf{138\;L}$$.