Given excess oxygen, how many moles of H_2O can be produced from 12.5 moles of C_4H_{10} in the complete combustion reaction: C_4H_{10} + O_2 ightarrow CO_2 + H_2O?

• A. 12.5 moles
• B. 50.0 moles
• C. 25.0 moles
• D. 37.5 moles

Answer: (D)

Verified step by step guidance

1. Write the balanced chemical equation for the complete combustion of butane (C_4H_{10}). The general form is: C_4H_{10} + O_2 \rightarrow CO_2 + H_2O.
2. Balance the carbon atoms first: since there are 4 carbons in C_4H_{10}, place a coefficient of 4 in front of CO_2: C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O.
3. Balance the hydrogen atoms next: there are 10 hydrogens in C_4H_{10}, so place a coefficient of 5 in front of H_2O (because each water molecule has 2 hydrogens): C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O.
4. Balance the oxygen atoms last: on the right side, there are (4 \times 2) + (5 \times 1) = 8 + 5 = 13 oxygen atoms. Since O_2 is diatomic, divide 13 by 2 to get the coefficient for O_2: C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4CO_2 + 5H_2O. To avoid fractions, multiply the entire equation by 2 if desired.
5. Use the mole ratio from the balanced equation between C_4H_{10} and H_2O to find the moles of water produced. For every 1 mole of C_4H_{10}, 5 moles of H_2O are produced. Multiply 12.5 moles of C_4H_{10} by 5 to get the moles of H_2O.