- Identify the stoichiometric ratio from the balanced chemical equation \(2 H_2 + O_2 \rightarrow 2 H_2 O\), which is \(2:1\) for \(H_2\) to \(O_2\).
- Use the given amount of \(H_2\) (\(27.4\) mol) to set up the ratio \(\frac{27.4 \text{ mol } H_2}{x \text{ mol } O_2} = \frac{2}{1}\).
- Solve for \(x\) to find the required moles of oxygen: \(x = \frac{27.4}{2} = 13.7\).
- The final answer is \(\boxed{13.7 \text{ mol}}\).
Explanation
- Problem Analysis and Stoichiometric Ratio
The problem asks for the amount of oxygen (\(O_2\)) required to react completely with \(27.4\) moles of hydrogen (\(H_2\)), based on the balanced chemical equation for the formation of water:
\[2 H_2 + O_2 \rightarrow 2 H_2 O\]
From the balanced equation, we can see that the stoichiometry of the reaction requires \(2\) moles of \(H_2\) for every \(1\) mole of \(O_2\). This gives us a molar ratio of \(\frac{1 \text{ mol } O_2}{2 \text{ mol } H_2}\).
- Calculation and Conclusion
We are given that we have \(27.4\) moles of \(H_2\). To find the moles of \(O_2\) required, we can use the ratio derived from the balanced equation:
\[x \text{ mol } O_2 = 27.4 \text{ mol } H_2 \times \left( \frac{1 \text{ mol } O_2}{2 \text{ mol } H_2} \right)\]
Performing the calculation:
\[x = \frac{27.4}{2} = 13.7 \text{ mol } O_2\]
Thus, \(13.7\) moles of oxygen are required to react with \(27.4\) moles of hydrogen. Comparing this to the given options:
A\(6.8\) mol
B\(13.7\) mol
C\(54.8\) mol
D\(109.6\) mol
The correct option is b.
Examples
Stoichiometry is a fundamental concept in chemistry used to determine the exact amounts of reactants needed to produce a specific amount of product. For example, a rocket engine needs to mix fuel (like hydrogen) and an oxidizer (like oxygen) in precise proportions to ensure efficient combustion. By using the balanced chemical equation, engineers can calculate the exact mass or moles of fuel and oxidizer required to power a space mission successfully.