Chemistrymodule 5504 Gas Calculations Worksheetcomplete The Calculati ✓ Solved

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 28oC? 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 25oC has its temperature increased to 35oC. 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 Scanned with CamScanner

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Gas Calculations Worksheet: Solutions

Gases follow several fundamental laws, notably Boyle’s Law, Charles’ Law, the Ideal Gas Law, and stoichiometry for gas reactions. The following calculations illustrate practical applications of these laws. The standard temperature and pressure (STP) is defined as 0°C (273.15 K) and 1 atm.

Problem 1: Volume of Nitrogen Gas at STP

Question: What is the volume of 2.5 moles of nitrogen gas (N₂) at standard temperature and pressure (STP)?
Solution:
At STP, 1 mole of an ideal gas occupies 22.4 liters. Thus, to find the volume:
\[
\text{Volume} = \text{Moles} \times 22.4\, \text{L/mol}
\]
\[
\text{Volume} = 2.5\, \text{moles} \times 22.4\, \text{L/mol} = 56.0\, \text{L}
\]

Problem 2: Liters of Water from Butane

Question: How many liters of water can be produced from 5.0 liters of butane gas at STP, assuming excess oxygen?
Balanced Reaction:
\[ C_4H_{10} + 13/2 O_2 \rightarrow 4 CO_2 + 5 H_2O \]
From the stoichiometry of the reaction, 1 mole of butane produces 5 moles of water. To convert liters to moles at STP:
\[
\text{Moles of Butane} = \frac{5.0\, \text{L}}{22.4\, \text{L/mol}} \approx 0.2232\, \text{moles}
\]
Using the 1:5 ratio:
\[
\text{Moles of Water} = 0.2232\, \text{moles} \times 5 = 1.116\, \text{moles}
\]
Converting moles of water back to liters:
\[
\text{Volume of Water} = 1.116\, \text{moles} \times 22.4\, \text{L/mol} \approx 25.0\, \text{L}
\]

Problem 3: Grams of Oxygen in a Sample

Question: How many grams of oxygen gas are contained in a 15 L sample at 1.02 atm and 28°C?
Solution:
First, convert temperature to Kelvin:
\[
T = 28 + 273.15 = 301.15\, K
\]
Using the Ideal Gas Law \( PV = nRT \):
\[
n = \frac{PV}{RT}
\]
Where
- \( R = 0.0821\, L\cdot atm/(K\cdot mol) \)
Substituting values:
\[
n = \frac{(1.02\, atm)(15\, L)}{(0.0821\, L\cdot atm/(K\cdot mol))(301.15\, K)} \approx 0.6\, \text{moles}
\]
Finding the mass of oxygen:
\[
\text{Mass} = n \cdot M = 0.6\, moles \cdot 32.00\, g/mol = 19.20\, g
\]

Problem 4: Liters of Oxygen Reacting with Iron

Question: How many liters of oxygen gas will react with 35.8 grams of iron?
Balanced Reaction:
\[ 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \]
First, convert grams of iron to moles:
\[
n_{Fe} = \frac{35.8\, g}{55.85\, g/mol} \approx 0.641\, moles
\]
According to the reaction, 4 moles of Fe react with 3 moles of O₂:
\[
\text{Moles of } O_2 = \frac{3}{4} \times 0.641 \approx 0.481\, moles
\]
Converting moles of O₂ to liters:
\[
\text{Volume} = 0.481\, moles \times 22.4\, L/mol \approx 10.77\, L
\]

Problem 5: Liters of Hydrogen from Sodium

Question: If 15.8 grams of sodium react with excess water, how many liters of hydrogen gas can be produced at 303 K and 1.30 atm?
Balanced Reaction:
\[ 2Na + 2H_2O \rightarrow 2NaOH + H_2 \]
Converting grams of sodium to moles:
\[
n_{Na} = \frac{15.8\, g}{22.99\, g/mol} \approx 0.688\, moles
\]
Hydrogen produced from sodium:
\[
\text{Moles of } H_2 = \frac{1}{2} n_{Na} = \frac{0.688}{2} \approx 0.344\, moles
\]
Calculate volume at specified conditions:
\[
V = \frac{nRT}{P} = \frac{(0.344\, moles)(0.0821\, L\cdot atm/(K\cdot mol))(303\, K)}{1.30\, atm} \approx 6.90\, L
\]

Problem 6: New Volume with Increased Pressure

Question: If 10.5 L of a gas at 0.98 atm has its pressure increased to 1.50 atm, what is the new volume?
Using Boyle's Law (\( P_1V_1 = P_2V_2 \)):
\[
(0.98\, atm)(10.5\, L) = (1.50\, atm)(V_2)
\]
Solving for \( V_2 \):
\[
V_2 = \frac{(0.98)(10.5)}{1.50} \approx 6.84\, L
\]

Problem 7: New Volume with Temperature Change

Question: 55 L of a gas at 25°C has its temperature increased to 35°C. What is its new volume?
Using Charles' Law (\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)):
Convert temperatures to Kelvin:
\[
T_1 = 25 + 273.15 = 298.15\, K; \quad T_2 = 35 + 273.15 = 308.15\, K
\]
\[
\frac{55\, L}{298.15\, K} = \frac{V_2}{308.15\, K}
\]
Solving for \( V_2 \):
\[
V_2 = 55\, L \times \frac{308.15\, K}{298.15\, K} \approx 57.5\, L
\]

Problem 8: New Volume from STP to Changed Conditions

Question: 158 L of a gas at STP has its conditions changed to 350 K and 1.50 atm. What is the new volume?
Using the Ideal Gas Law and conversion:
\[
V = \frac{P_1V_1T_2}{P_2T_1}
\]
Where:
- \( P_1 = 1 atm, V_1 = 158 L, T_1 = 273.15\, K, P_2 = 1.50\, atm, T_2 = 350 K \)
\[
V = \frac{(1)(158)(350)}{(1.50)(273.15)} \approx 98.77\, L
\]

References

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This worksheet provides systematic applications of gas laws through a variety of stoichiometric and theoretical calculations, demonstrating the versatility and utility of these principles in chemistry.