Question

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)

Hint:

Always balance the equation first. For combustion of alkanes \(C_nH_{2n+2}\), the moles of water formed is always \(n+1\).

Answer 1

Correct Answer: 464

Step 1: Understanding the Concept:
Stoichiometry relates the quantities of reactants and products in a balanced chemical reaction.
We use the molar mass to convert between mass and moles.
Step 2: Key Formula or Approach:
1. Balanced Reaction: \(C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O\).
2. Mole-mass calculation: \(m = n \times M\).
Step 3: Detailed Explanation:
1. Calculate moles of water produced:
Molar mass of \(H_2O = 18 \text{ g/mol}\).
\[ n_{H_2O} = \frac{72.0}{18} = 4.0 \text{ mol} \]

2. Calculate moles of butane (\(C_4H_{10}\)) needed:
From balanced equation, 1 mole of butane produces 5 moles of water.
\[ n_{C_4H_{10}} = \frac{1}{5} \times n_{H_2O} = \frac{4.0}{5} = 0.8 \text{ mol} \]

3. Calculate mass of butane:
Molar mass of \(C_4H_{10} = 4(12) + 10(1) = 58 \text{ g/mol}\).
\[ \text{Mass} = 0.8 \times 58 = 46.4 \text{ g} \]
Expressing in \(10^{-1}\) g format: \(464 \times 10^{-1}\) g.
Step 4: Final Answer:
The value of \(x\) is 464.