Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation: 2Al(s) + 2NAOH(aq) + 6H20(1)-2NAAI(OH)4(aq) + 3H2(g) The product gas, H2, is collected over water at a temperature of 25 °C and a pressure of 744 mm Hg. If the wet H, gas formed occupies a volume of 6.55 L, the number of moles of Al reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.
Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation: 2Al(s) + 2NAOH(aq) + 6H20(1)-2NAAI(OH)4(aq) + 3H2(g) The product gas, H2, is collected over water at a temperature of 25 °C and a pressure of 744 mm Hg. If the wet H, gas formed occupies a volume of 6.55 L, the number of moles of Al reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.
Chemistry
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Chapter1: Chemical Foundations
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![**Chemical Reaction of Aluminum with Sodium Hydroxide to Produce Hydrogen Gas**
Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation:
\[ 2\text{Al(s)} + 2\text{NaOH(aq)} + 6\text{H}_2\text{O(l)} \rightarrow 2\text{NaAl(OH)}_4\text{(aq)} + 3\text{H}_2\text{(g)} \]
In this reaction, the product gas, \(\text{H}_2\), is collected over water at a temperature of 25°C and a pressure of 744 mm Hg. If the wet \(\text{H}_2\) gas formed occupies a volume of 6.55 L, the number of moles of Al reacted was ____ mol. The vapor pressure of water is 23.8 mm Hg at 25°C.
**Explanation:**
- **Reaction Description**: Aluminum (Al) reacts with sodium hydroxide (NaOH) and water (H₂O) to produce sodium aluminate (\(\text{NaAl(OH)}_4\)) in aqueous solution and hydrogen gas (\(\text{H}_2\)).
- **Conditions**: The reaction conditions include a temperature of 25°C and a pressure of 744 mm Hg. The vapor pressure of water at this temperature is 23.8 mm Hg.
- **Objective**: Calculate the number of moles of aluminum that reacted, given that the volume of the gas collected is 6.55 L.
Graphs or diagrams were not present in the image.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F924f1927-2b95-447c-8315-cf80b16c3d9e%2F32031a30-31a7-4abb-beed-5f887eab6fe1%2Fttgm043_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction of Aluminum with Sodium Hydroxide to Produce Hydrogen Gas**
Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation:
\[ 2\text{Al(s)} + 2\text{NaOH(aq)} + 6\text{H}_2\text{O(l)} \rightarrow 2\text{NaAl(OH)}_4\text{(aq)} + 3\text{H}_2\text{(g)} \]
In this reaction, the product gas, \(\text{H}_2\), is collected over water at a temperature of 25°C and a pressure of 744 mm Hg. If the wet \(\text{H}_2\) gas formed occupies a volume of 6.55 L, the number of moles of Al reacted was ____ mol. The vapor pressure of water is 23.8 mm Hg at 25°C.
**Explanation:**
- **Reaction Description**: Aluminum (Al) reacts with sodium hydroxide (NaOH) and water (H₂O) to produce sodium aluminate (\(\text{NaAl(OH)}_4\)) in aqueous solution and hydrogen gas (\(\text{H}_2\)).
- **Conditions**: The reaction conditions include a temperature of 25°C and a pressure of 744 mm Hg. The vapor pressure of water at this temperature is 23.8 mm Hg.
- **Objective**: Calculate the number of moles of aluminum that reacted, given that the volume of the gas collected is 6.55 L.
Graphs or diagrams were not present in the image.
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