Answer :
To find the volume of hydrogen gas produced from the reaction of 38.8 grams of aluminum (Al) with excess hydrochloric acid (HCl) at standard temperature and pressure (STP), we can follow these steps:
1. Determine the Moles of Aluminum:
- First, we need to know the molar mass of aluminum (Al), which is approximately 26.98 grams per mole.
- Using the molar mass, we calculate the moles of aluminum:
[tex]\[
\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{38.8 \, \text{g}}{26.98 \, \text{g/mol}}
\][/tex]
- This calculation gives us approximately 1.44 moles of aluminum.
2. Use Stoichiometry to Find Moles of Hydrogen Gas (H₂) Produced:
- According to the balanced chemical equation:
[tex]\[
2 \, \text{Al} + 6 \, \text{HCl} \rightarrow 2 \, \text{AlCl}_3 + 3 \, \text{H}_2
\][/tex]
- From this equation, 2 moles of aluminum produce 3 moles of hydrogen gas.
- Therefore, the moles of hydrogen gas produced are:
[tex]\[
\text{Moles of H₂} = \left(\frac{3}{2}\right) \times \text{Moles of Al} = \left(\frac{3}{2}\right) \times 1.44
\][/tex]
- This calculation results in approximately 2.16 moles of hydrogen gas.
3. Calculate the Volume of Hydrogen Gas at STP:
- At STP, 1 mole of any ideal gas occupies approximately 22.4 liters.
- Therefore, the volume of hydrogen gas produced is:
[tex]\[
\text{Volume of H₂} = \text{Moles of H₂} \times 22.4 \, \text{L/mol} = 2.16 \times 22.4
\][/tex]
- This results in a volume of approximately 48.32 liters of hydrogen gas.
In summary, the reaction of 38.8 grams of aluminum with excess hydrochloric acid at STP produces approximately 48.32 liters of hydrogen gas.
1. Determine the Moles of Aluminum:
- First, we need to know the molar mass of aluminum (Al), which is approximately 26.98 grams per mole.
- Using the molar mass, we calculate the moles of aluminum:
[tex]\[
\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{38.8 \, \text{g}}{26.98 \, \text{g/mol}}
\][/tex]
- This calculation gives us approximately 1.44 moles of aluminum.
2. Use Stoichiometry to Find Moles of Hydrogen Gas (H₂) Produced:
- According to the balanced chemical equation:
[tex]\[
2 \, \text{Al} + 6 \, \text{HCl} \rightarrow 2 \, \text{AlCl}_3 + 3 \, \text{H}_2
\][/tex]
- From this equation, 2 moles of aluminum produce 3 moles of hydrogen gas.
- Therefore, the moles of hydrogen gas produced are:
[tex]\[
\text{Moles of H₂} = \left(\frac{3}{2}\right) \times \text{Moles of Al} = \left(\frac{3}{2}\right) \times 1.44
\][/tex]
- This calculation results in approximately 2.16 moles of hydrogen gas.
3. Calculate the Volume of Hydrogen Gas at STP:
- At STP, 1 mole of any ideal gas occupies approximately 22.4 liters.
- Therefore, the volume of hydrogen gas produced is:
[tex]\[
\text{Volume of H₂} = \text{Moles of H₂} \times 22.4 \, \text{L/mol} = 2.16 \times 22.4
\][/tex]
- This results in a volume of approximately 48.32 liters of hydrogen gas.
In summary, the reaction of 38.8 grams of aluminum with excess hydrochloric acid at STP produces approximately 48.32 liters of hydrogen gas.
