Cva Chemistry Module 5504 Gas Calculations Worksheet
Cva Chemistrymodule 5504 Gas Calculations Worksheetcomplete The Calcu
CVA Chemistry Module .04 Gas Calculations Worksheet Complete the calculations below, showing all your work. To practice gas law calculations before taking the quiz. 1. What is the volume of 2.5 moles of nitrogen gas (N2) at standard temperature and pressure (STP)? 2. How many liters of water can be produced from 5.0 liters of butane gas at STP, assuming excess oxygen? C4H10 (g) + O2 (g) CO2 (g) + H2O (g) 3. How many grams of oxygen gas are contained in a 15 L sample at 1.02 atm and 28°C? 4. How many liters of oxygen gas, at standard temperature and pressure, will react with 35.8 grams of iron metal? 4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) 5. If 15.8 grams of sodium react with excess water, how many liters of hydrogen gas can be produced at 303 Kelvin and 1.30 atmospheres? 2 Na (s) + 2 H2O 2 NaOH (l) + H2 (g) 6. If 10.5 L of a gas at 0.98 atm has its pressure increased to 1.50 atm, what is the new volume? 7. 55 L of a gas at 25°C has its temperature increased to 35°C. What is its new volume? 8. 158 L of a gas at STP has its conditions changed to 350 K and 1.50 atm. What is the new volume? Save and submit to 5.04 Gas Calculations Worksheet
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Introduction
Gas laws are fundamental in understanding the behavior of gases under different conditions of temperature, pressure, and volume. The application of these laws allows chemists to predict how gases will respond to changes, which is critical in both laboratory and industrial settings. This paper addresses several typical gas law problems, demonstrating calculations based on the ideal gas law, combined gas law, and other related principles to elucidate the relationships among gas variables.
Problem 1: Volume of Nitrogen Gas at STP
Given 2.5 moles of nitrogen gas (N₂) at standard temperature and pressure (STP), the goal is to find its volume. STP is defined as a temperature of 273.15 K (0°C) and a pressure of 1 atm. According to Avogadro's law, at STP, one mole of any ideal gas occupies 22.4 liters. Therefore, the volume of 2.5 moles of N₂ can be calculated as:
\[ V = n \times 22.4\, \text{L} = 2.5 \times 22.4 = 56\, \text{L} \]
Thus, the volume of 2.5 moles of nitrogen gas at STP is 56 liters.
Problem 2: Water Production from Butane Gas
The reaction between butane (C₄H₁₀) and oxygen results in carbon dioxide and water:
\[ C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \]
The molar ratio indicates 1 mol of butane produces 1 mol of water. At STP, 5.0 liters of butane corresponds to moles:
\[ n = \frac{V}{22.4\, \text{L}} = \frac{5.0}{22.4} \approx 0.223 \,\text{mol} \]
Thus, 0.223 mol of water can be produced, which at STP occupies:
\[ V_{H_2O} = 0.223 \times 22.4 \approx 5.0\, \text{L} \]
Therefore, approximately 5.0 liters of water can be generated from 5.0 liters of butane gas under these conditions.
Problem 3: Mass of Oxygen in a Gas Sample
Using the ideal gas law \( PV = nRT \), we find the number of moles of oxygen:
\[ n = \frac{PV}{RT} \]
where \( P = 1.02\, \text{atm} \), \( V = 15\, \text{L} \), \( R = 0.0821\, \text{L·atm/(mol·K)} \), and \( T = 28^\circ C = 301\, \text{K} \).
\[ n = \frac{1.02 \times 15}{0.0821 \times 301} \approx 0.616\, \text{mol} \]
The mass is:
\[ m = n \times M_{O_2} = 0.616 \times 32 = 19.7\, \text{g} \]
Hence, the sample contains approximately 19.7 grams of oxygen gas.
Problem 4: Volume of Oxygen Reacting with Iron
The reaction:
\[ 4\, \text{Fe} + 3\, O_2 \rightarrow 2\, Fe_2O_3 \]
moles of iron:
\[ n_{Fe} = \frac{35.8}{55.85} \approx 0.642\, \text{mol} \]
From the stoichiometry, \( 4\, \text{mol}\, Fe \) reacts with \( 3\, \text{mol}\, O_2 \), thus:
\[ n_{O_2} = \frac{3}{4} \times n_{Fe} \approx 0.482\, \text{mol} \]
Using the ideal gas law to find the volume at STP:
\[ V = n \times 22.4 = 0.482 \times 22.4 \approx 10.8\, \text{L} \]
Thus, approximately 10.8 liters of oxygen are required.
Problem 5: Hydrogen Gas Production from Sodium
From:
\[ 2\, Na + 2\, H_2O \rightarrow 2\, NaOH + H_2 \]
moles of sodium:
\[ n_{Na} = \frac{15.8}{22.99} \approx 0.688\, \text{mol} \]
Hydrogen produced:
\[ n_{H_2} = 0.688\, \text{mol} \]
At 303 K and 1.30 atm, and using the ideal gas law:
\[ V = \frac{nRT}{P} = \frac{0.688 \times 0.0821 \times 303}{1.30} \approx 13.0\, \text{L} \]
Therefore, approximately 13.0 liters of hydrogen gas can be produced.
Problem 6: Change in Gas Volume with Pressure
Applying Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
where \( P_1 = 0.98\, \text{atm} \), \( V_1 = 10.5\, \text{L} \), \( P_2 = 1.50\, \text{atm} \):
\[ V_2 = \frac{P_1 V_1}{P_2} = \frac{0.98 \times 10.5}{1.50} \approx 6.85\, \text{L} \]
The new volume is approximately 6.85 liters.
Problem 7: Effect of Temperature Increase on Gas Volume
Using Charles's Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
convert the temperatures to Kelvin:
\[ T_1 = 25^\circ C = 298\, \text{K} \]
\[ T_2 = 35^\circ C = 308\, \text{K} \]
Calculate:
\[ V_2 = V_1 \times \frac{T_2}{T_1} = 55 \times \frac{308}{298} \approx 56.9\, \text{L} \]
So, the volume increases to approximately 56.9 liters.
Problem 8: Change in Volume with Temperature and Pressure
Using combined gas law:
\[ \frac{V_1 P_1}{T_1} = \frac{V_2 P_2}{T_2} \]
At STP:
\[ V_1 = 158\, \text{L},\, P_1 = 1\, \text{atm},\, T_1 = 273\, \text{K} \]
New conditions:
\[ P_2 = 1.50\, \text{atm},\, T_2 = 350\, \text{K} \]
Calculate \( V_2 \):
\[ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 158 \times \frac{1}{1.50} \times \frac{350}{273} \approx 158 \times 0.6667 \times 1.282 \approx 134.9\, \text{L} \]
The new volume is approximately 134.9 liters.
Conclusion
This set of gas law calculations demonstrates the practical application of fundamental principles such as Avogadro's law, Boyle's law, Charles's law, and the ideal gas law. Mastery of these calculations enables chemists to predict gas behaviors under varying conditions, essential in laboratory research, industrial processes, and environmental science. Accurate understanding and application of these principles foster better scientific predictions and technological advancements.
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