Answer :
Sure! Let's solve the problem step-by-step.
### Step 1: Understand the Chemical Reaction
The reaction given involves aluminum (Al) reacting with sodium hydroxide (NaOH) to produce hydrogen gas (H₂) and sodium aluminate (Na₃AlO₃). The balanced chemical equation is:
[tex]\[ 2\text{Al} + 6\text{NaOH} \rightarrow 2\text{Na}_3\text{AlO}_3 + 3\text{H}_2 \][/tex]
### Step 2: Use Ideal Gas Law to Find Moles of Hydrogen
We need to find how many moles of hydrogen gas are produced. We'll use the Ideal Gas Law:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the universal gas constant ([tex]\( 0.0821 \, \text{L atm/mol K} \)[/tex]),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
Given data:
- Pressure [tex]\( P = 98.5 \, \text{kPa} \)[/tex]. First, convert this to atm: [tex]\( P = 98.5 \, \text{kPa} \times \frac{1 \, \text{atm}}{101.325 \, \text{kPa}} \approx 0.972 \, \text{atm} \)[/tex].
- Volume [tex]\( V = 5.00 \, \text{L} \)[/tex].
- Temperature [tex]\( T = 15^{\circ}\text{C} = 15 + 273.15 = 288.15 \, \text{K} \)[/tex].
Plug these values into the Ideal Gas Law to find [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} = \frac{(0.972 \, \text{atm})(5.00 \, \text{L})}{(0.0821 \, \text{L atm/mol K})(288.15 \, \text{K})} \approx 0.205 \, \text{mol} \][/tex]
### Step 3: Use Stoichiometry to Find Moles of Aluminum
From the balanced equation, we see that 2 moles of aluminum produce 3 moles of hydrogen gas. We can set up a ratio to find moles of aluminum needed:
[tex]\[
\frac{2 \, \text{moles of Al}}{3 \, \text{moles of H}_2} \times 0.205 \, \text{mol H}_2 \approx 0.137 \, \text{mol Al}
\][/tex]
### Step 4: Convert Moles of Aluminum to Grams
Finally, use the molar mass of aluminum (Al), which is [tex]\( 26.98 \, \text{g/mol} \)[/tex], to convert moles to grams:
[tex]\[
0.137 \, \text{mol Al} \times 26.98 \, \text{g/mol} \approx 3.70 \, \text{g Al}
\][/tex]
Therefore, approximately 3.70 grams of aluminum are needed to produce 5.00 liters of hydrogen gas under the given conditions.
### Step 1: Understand the Chemical Reaction
The reaction given involves aluminum (Al) reacting with sodium hydroxide (NaOH) to produce hydrogen gas (H₂) and sodium aluminate (Na₃AlO₃). The balanced chemical equation is:
[tex]\[ 2\text{Al} + 6\text{NaOH} \rightarrow 2\text{Na}_3\text{AlO}_3 + 3\text{H}_2 \][/tex]
### Step 2: Use Ideal Gas Law to Find Moles of Hydrogen
We need to find how many moles of hydrogen gas are produced. We'll use the Ideal Gas Law:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the universal gas constant ([tex]\( 0.0821 \, \text{L atm/mol K} \)[/tex]),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
Given data:
- Pressure [tex]\( P = 98.5 \, \text{kPa} \)[/tex]. First, convert this to atm: [tex]\( P = 98.5 \, \text{kPa} \times \frac{1 \, \text{atm}}{101.325 \, \text{kPa}} \approx 0.972 \, \text{atm} \)[/tex].
- Volume [tex]\( V = 5.00 \, \text{L} \)[/tex].
- Temperature [tex]\( T = 15^{\circ}\text{C} = 15 + 273.15 = 288.15 \, \text{K} \)[/tex].
Plug these values into the Ideal Gas Law to find [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} = \frac{(0.972 \, \text{atm})(5.00 \, \text{L})}{(0.0821 \, \text{L atm/mol K})(288.15 \, \text{K})} \approx 0.205 \, \text{mol} \][/tex]
### Step 3: Use Stoichiometry to Find Moles of Aluminum
From the balanced equation, we see that 2 moles of aluminum produce 3 moles of hydrogen gas. We can set up a ratio to find moles of aluminum needed:
[tex]\[
\frac{2 \, \text{moles of Al}}{3 \, \text{moles of H}_2} \times 0.205 \, \text{mol H}_2 \approx 0.137 \, \text{mol Al}
\][/tex]
### Step 4: Convert Moles of Aluminum to Grams
Finally, use the molar mass of aluminum (Al), which is [tex]\( 26.98 \, \text{g/mol} \)[/tex], to convert moles to grams:
[tex]\[
0.137 \, \text{mol Al} \times 26.98 \, \text{g/mol} \approx 3.70 \, \text{g Al}
\][/tex]
Therefore, approximately 3.70 grams of aluminum are needed to produce 5.00 liters of hydrogen gas under the given conditions.
