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To determine the volume of carbon dioxide (CO[tex]\(_2\)[/tex]) that can be removed from the cabin air using 1.00 grams of sodium hydroxide (NaOH), we need to follow a step-by-step approach based on stoichiometry and the conditions of standard temperature and pressure (STP).
### Step 1: Determine the Number of Moles of NaOH
First, we need to calculate the number of moles of NaOH from its mass. We use the molar mass of NaOH, which is given as 40.00 grams per mole.
Given:
- Mass of NaOH = 1.00 grams
- Molar mass of NaOH = 40.00 grams/mole
The number of moles of NaOH can be found using the formula:
[tex]\[ \text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of NaOH} = \frac{1.00 \, \text{grams}}{40.00 \, \text{grams/mole}} = 0.025 \, \text{moles} \][/tex]
### Step 2: Determine the Number of Moles of CO[tex]\(_2\)[/tex]
From the balanced chemical equation:
[tex]\[ 2 \text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \][/tex]
We see that 2 moles of NaOH react with 1 mole of CO[tex]\(_2\)[/tex]. Therefore, the ratio of moles of NaOH to moles of CO[tex]\(_2\)[/tex] is 2:1.
Using this ratio to find the number of moles of CO[tex]\(_2\)[/tex]:
[tex]\[ \text{moles of CO}_2 = \frac{\text{moles of NaOH}}{2} \][/tex]
For the calculated moles of NaOH (0.025 moles):
[tex]\[ \text{moles of CO}_2 = \frac{0.025 \, \text{moles}}{2} = 0.0125 \, \text{moles} \][/tex]
### Step 3: Calculate the Volume of CO[tex]\(_2\)[/tex] at STP
Standard Temperature and Pressure (STP) conditions define the molar volume of a gas as 22.4 liters per mole. We use this molar volume to calculate the volume of CO[tex]\(_2\)[/tex].
Given:
- Molar volume of CO[tex]\(_2\)[/tex] at STP = 22.4 liters/mole
The volume of CO[tex]\(_2\)[/tex] can be calculated using the formula:
[tex]\[ \text{volume of CO}_2 = \text{moles of CO}_2 \times \text{molar volume} \][/tex]
Plugging in the values:
[tex]\[ \text{volume of CO}_2 = 0.0125 \, \text{moles} \times 22.4 \, \text{liters/mole} = 0.280 \, \text{liters} \][/tex]
Therefore, the volume of carbon dioxide that can be removed from the cabin air with 1.00 grams of sodium hydroxide is 0.280 liters.
### Step 1: Determine the Number of Moles of NaOH
First, we need to calculate the number of moles of NaOH from its mass. We use the molar mass of NaOH, which is given as 40.00 grams per mole.
Given:
- Mass of NaOH = 1.00 grams
- Molar mass of NaOH = 40.00 grams/mole
The number of moles of NaOH can be found using the formula:
[tex]\[ \text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of NaOH} = \frac{1.00 \, \text{grams}}{40.00 \, \text{grams/mole}} = 0.025 \, \text{moles} \][/tex]
### Step 2: Determine the Number of Moles of CO[tex]\(_2\)[/tex]
From the balanced chemical equation:
[tex]\[ 2 \text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \][/tex]
We see that 2 moles of NaOH react with 1 mole of CO[tex]\(_2\)[/tex]. Therefore, the ratio of moles of NaOH to moles of CO[tex]\(_2\)[/tex] is 2:1.
Using this ratio to find the number of moles of CO[tex]\(_2\)[/tex]:
[tex]\[ \text{moles of CO}_2 = \frac{\text{moles of NaOH}}{2} \][/tex]
For the calculated moles of NaOH (0.025 moles):
[tex]\[ \text{moles of CO}_2 = \frac{0.025 \, \text{moles}}{2} = 0.0125 \, \text{moles} \][/tex]
### Step 3: Calculate the Volume of CO[tex]\(_2\)[/tex] at STP
Standard Temperature and Pressure (STP) conditions define the molar volume of a gas as 22.4 liters per mole. We use this molar volume to calculate the volume of CO[tex]\(_2\)[/tex].
Given:
- Molar volume of CO[tex]\(_2\)[/tex] at STP = 22.4 liters/mole
The volume of CO[tex]\(_2\)[/tex] can be calculated using the formula:
[tex]\[ \text{volume of CO}_2 = \text{moles of CO}_2 \times \text{molar volume} \][/tex]
Plugging in the values:
[tex]\[ \text{volume of CO}_2 = 0.0125 \, \text{moles} \times 22.4 \, \text{liters/mole} = 0.280 \, \text{liters} \][/tex]
Therefore, the volume of carbon dioxide that can be removed from the cabin air with 1.00 grams of sodium hydroxide is 0.280 liters.
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