Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ___ $$\times 10^{-1}$$ g. (in nearest integer)

For the complete combustion of butane, we first recall and write the balanced chemical equation:

$$2\,\mathrm{C_4H_{10}} \;+\; 13\,\mathrm{O_2}\;\longrightarrow\; 8\,\mathrm{CO_2}\;+\; 10\,\mathrm{H_2O}$$

From this equation we see that:

$$2 \text{ mol } \mathrm{C_4H_{10}} \;\xrightarrow{\text{combustion}}\; 10 \text{ mol } \mathrm{H_2O}$$

Now, we are told that the actual amount of water obtained is $$72.0\ \text{g}$$. Let us convert this mass of water into moles. The molar mass of water $$\mathrm{H_2O}$$ is

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

Therefore, the number of moles of water produced is

$$n(\mathrm{H_2O}) \;=\;\frac{72.0\ \text{g}}{18\ \text{g mol}^{-1}} \;=\; 4.0\ \text{mol}.$$

According to the stoichiometric ratio obtained from the balanced equation, $$2$$ moles of butane give $$10$$ moles of water. Hence, to find the moles of butane required for $$4.0$$ moles of water we write the proportion

$$\frac{2\ \text{mol }\mathrm{C_4H_{10}}}{10\ \text{mol }\mathrm{H_2O}} \;=\;\frac{x\ \text{mol }\mathrm{C_4H_{10}}}{4.0\ \text{mol }\mathrm{H_2O}}.$$

Cross-multiplying gives

$$10\,x = 2 \times 4.0,$$

so

$$x = \frac{2 \times 4.0}{10} = 0.8\ \text{mol}.$$

Now we convert the required moles of butane to mass. The molar mass of butane ($$\mathrm{C_4H_{10}}$$) is

$$M(\mathrm{C_4H_{10}}) = (4 \times 12) + (10 \times 1) = 48 + 10 = 58\ \text{g mol}^{-1}.$$

Hence, the mass of butane consumed is

$$m(\mathrm{C_4H_{10}}) = n \times M = 0.8\ \text{mol} \times 58\ \text{g mol}^{-1} = 46.4\ \text{g}.$$

The question asks us to express this answer in the form $$\_\_\_\_\_\times 10^{-1}\ \text{g}$$ and then give the blank as the nearest integer. We rewrite

$$46.4\ \text{g} = 464 \times 10^{-1}\ \text{g}.$$

So the integer that fills the blank is $$464$$.

Hence, the correct answer is Option 464.