Express your answer to two significant figures.
You form water vapor by mixing oxygen and hydrogen at $730^{\circ} C$ in a 5.4-liter container. This is the equation for the reaction:
[tex]$O_2(g)+2 H_2(g) \rightarrow 2 H_2 O(g)$[/tex]
The partial pressure of oxygen before the reaction is 122.3 kilopascals, and there is excess hydrogen. How many moles of water are formed?
The reaction produces $\square$ moles of water.
Answer (2)
Convert the temperature from Celsius to Kelvin: T ( K ) = 730 + 273.15 = 1003.15 K .
Convert the pressure from kPa to Pa: P ( P a ) = 122.3 × 1000 = 122300 P a .
Use the ideal gas law to find the moles of oxygen: n O 2 = 8.314 × 1003.15 122300 × 5.4 × 1 0 − 3 ≈ 0.079 m o l .
Calculate the moles of water formed: n H 2 O = 2 × n O 2 = 2 × 0.079 ≈ 0.16 m o l . The reaction produces 0.16 moles of water.
Explanation
Convert Celsius to Kelvin First, we need to convert the temperature from Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperature: T ( K ) = T ( ∘ C ) + 273.15 = 730 + 273.15 = 1003.15 K .
Convert kPa to Pa Next, we convert the pressure from kilopascals (kPa) to pascals (Pa) by multiplying by 1000: P ( P a ) = P ( k P a ) × 1000 = 122.3 × 1000 = 122300 P a .
Calculate moles of Oxygen Now, we use the ideal gas law to find the number of moles of oxygen ( n O 2 ). The ideal gas law is given by P V = n RT , where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We have P = 122300 P a , V = 5.4 L = 0.0054 m 3 , R = 8.314 J / ( m o l ⋅ K ) , and T = 1003.15 K . Rearranging the ideal gas law to solve for n , we get n = RT P V . Substituting the values, we have n O 2 = 8.314 × 1003.15 122300 × 5.4 × 1 0 − 3 = 8340.55 650.42 ≈ 0.078
Calculate moles of Water From the balanced chemical equation, O 2 ( g ) + 2 H 2 ( g ) → 2 H 2 O ( g ) , the mole ratio of O 2 to H 2 O is 1:2. Therefore, the number of moles of water formed is twice the number of moles of oxygen initially present: n H 2 O = 2 × n O 2 = 2 × 0.078 = 0.156 .
Round to significant figures Finally, we round the number of moles of water to two significant figures, which gives us 0.16 moles.
Examples
Understanding stoichiometry and gas laws is crucial in various real-world applications. For instance, in the Haber-Bosch process, nitrogen and hydrogen react to form ammonia, a key component of fertilizers. By carefully controlling the pressure, temperature, and reactant ratios, engineers can optimize ammonia production. Similarly, in internal combustion engines, precise air-fuel mixtures are essential for efficient combustion and minimizing emissions. The ideal gas law and stoichiometric principles help engineers design and operate these systems effectively, ensuring optimal performance and resource utilization. These principles are also vital in designing airbags, where a rapid chemical reaction produces gas to inflate the bag and protect vehicle occupants during a collision. The amount of gas produced must be carefully calculated to ensure proper inflation.
The reaction produces approximately 0.16 moles of water vapor by mixing oxygen and hydrogen at 730°C in a 5.4-liter container. This calculation involved using the ideal gas law and converting units for accurate results. It is based on the stoichiometric relationship from the balanced chemical equation.
;
