Gas Stoichiometry Worksheet Name: Solve all the following gas law problems. Show all work, answers are given at the end of the problem. Molar Volume 1. Calculate the number of moles contained in 550.mL of carbon dioxide at STP. (0.0246mol) 2. Calculate the mass of 1.50 L of CH4 at STP. (1.07g) 3. Calculate the volume in liters of 50.0 grams of nitrogen dioxide at STP. (24.3L) Gas Stoichiometry 4. Given the following reaction: NH.NO, (s) + N2(g) + 2H20 (g) How many liters of nitrogen gas is produced if 50.0L of water is produced at STP? (25.0L) 5. Given the following reaction: CuO (s) + H2(g) ? Cu (s) + H20 (g) If 250. L of hydrogen gas are used to reduce copper (II) oxide at STP, what mass of copper is to be expected? (7095) 6. How many grams of solid lithium must be added to liquid water in order to obtain 15.0 L of hydrogen gas at STP? (9.29g) 2 Li + 21,0 + 2LiOH + H2
The Correct Answer and Explanation is :
Let’s go through each of the gas law problems step by step, showing the work and explaining the calculations.
1. Number of moles in 550 mL of CO₂ at STP
We use the molar volume at STP (Standard Temperature and Pressure) where 1 mole of gas occupies 22.4 L (or 22.4 × 10³ mL).
Given:
- Volume of CO₂ = 550 mL = 0.550 L
- Molar Volume at STP = 22.4 L/mol
Formula:
[
\text{Moles of gas} = \frac{\text{Volume of gas}}{\text{Molar Volume}}
]
[
\text{Moles of CO₂} = \frac{0.550}{22.4} = 0.0246 \, \text{mol}
]
2. Mass of 1.50 L of CH₄ at STP
To calculate the mass, we need the molar mass of CH₄, which is 16.04 g/mol. Using the molar volume at STP:
Given:
- Volume of CH₄ = 1.50 L
- Molar Volume at STP = 22.4 L/mol
- Molar Mass of CH₄ = 16.04 g/mol
Steps:
- Calculate moles of CH₄:
[
\text{Moles of CH₄} = \frac{1.50}{22.4} = 0.0670 \, \text{mol}
] - Now, calculate the mass of CH₄:
[
\text{Mass} = 0.0670 \times 16.04 = 1.07 \, \text{g}
]
3. Volume of 50.0 grams of NO₂ at STP
To solve this, first, find the moles of NO₂, then use the molar volume to calculate the volume.
Given:
- Mass of NO₂ = 50.0 g
- Molar Mass of NO₂ = 46.0 g/mol
- Molar Volume at STP = 22.4 L/mol
Steps:
- Calculate moles of NO₂:
[
\text{Moles of NO₂} = \frac{50.0}{46.0} = 1.087 \, \text{mol}
] - Now, calculate the volume of NO₂:
[
\text{Volume} = 1.087 \times 22.4 = 24.3 \, \text{L}
]
4. Liters of N₂ produced if 50.0 L of H₂O is produced
The reaction is:
[
\text{NH₄NO₃ (s)} \rightarrow \text{N₂(g)} + 2 \text{H₂O (g)}
]
From the balanced equation, 1 mole of NH₄NO₃ produces 1 mole of N₂ and 2 moles of H₂O. If 50.0 L of H₂O is produced, we can use the molar volume at STP to relate volumes of gases.
Given:
- Volume of H₂O = 50.0 L
From the balanced equation:
- 2 L of H₂O is produced for every 1 L of N₂. So, if 50.0 L of H₂O is produced, the volume of N₂ produced will be half of that.
[
\text{Volume of N₂} = \frac{50.0}{2} = 25.0 \, \text{L}
]
5. Mass of Copper produced from 250 L of H₂ at STP
The reaction is:
[
\text{CuO (s)} + \text{H₂ (g)} \rightarrow \text{Cu (s)} + \text{H₂O (g)}
]
From the balanced equation, 1 mole of H₂ reacts to form 1 mole of Cu. We can use the molar volume to find the moles of H₂ and then the mass of Cu.
Given:
- Volume of H₂ = 250. L
- Molar Volume = 22.4 L/mol
- Molar Mass of Cu = 63.5 g/mol
Steps:
- Calculate moles of H₂:
[
\text{Moles of H₂} = \frac{250}{22.4} = 11.16 \, \text{mol}
] - Since 1 mole of H₂ produces 1 mole of Cu, the moles of Cu produced will be the same as the moles of H₂:
[
\text{Moles of Cu} = 11.16 \, \text{mol}
] - Now, calculate the mass of Cu:
[
\text{Mass of Cu} = 11.16 \times 63.5 = 709.5 \, \text{g}
]
6. Grams of Lithium needed to obtain 15.0 L of H₂ at STP
The reaction is:
[
2 \text{Li} + 2 \text{H₂O} \rightarrow 2 \text{LiOH} + \text{H₂}
]
From the balanced equation, 2 moles of Li produce 1 mole of H₂. We can use the molar volume to find moles of H₂ and then calculate the mass of lithium required.
Given:
- Volume of H₂ = 15.0 L
- Molar Volume = 22.4 L/mol
- Molar Mass of Li = 6.94 g/mol
Steps:
- Calculate moles of H₂:
[
\text{Moles of H₂} = \frac{15.0}{22.4} = 0.669 \, \text{mol}
] - From the equation, 2 moles of Li produce 1 mole of H₂, so moles of Li required are twice the moles of H₂:
[
\text{Moles of Li} = 0.669 \times 2 = 1.338 \, \text{mol}
] - Now, calculate the mass of Li:
[
\text{Mass of Li} = 1.338 \times 6.94 = 9.29 \, \text{g}
]
Summary of answers:
- 0.0246 mol
- 1.07 g
- 24.3 L
- 25.0 L of N₂
- 709.5 g of Cu
- 9.29 g of Li