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Answer :
Sure! Let's go through the process of finding out how many grams of aluminum react in this chemical reaction step-by-step. We will follow these steps:
1. Understand the Chemical Reaction:
- The reaction between aluminum (Al) and hydrochloric acid (HCl) produces aluminum chloride (AlCl₃) and hydrogen gas (H₂).
- The balanced equation is:
[tex]\[
2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2
\][/tex]
- From the equation, we can see that 2 moles of aluminum produce 3 moles of hydrogen gas.
2. Use the Ideal Gas Law to Find Moles of Hydrogen Gas:
- Given:
- Volume of hydrogen gas = 35.8 mL
- Temperature = [tex]\(27^\circ C\)[/tex]
- Pressure = 751 mmHg
- First, convert the temperature to Kelvin:
[tex]\[
\text{Temperature in Kelvin} = 27 + 273.15 = 300.15 \, \text{K}
\][/tex]
- Convert the pressure from mmHg to atm:
[tex]\[
\text{Pressure in atm} = \frac{751 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.9881 \, \text{atm}
\][/tex]
- Convert the volume from mL to L:
[tex]\[
\text{Volume in liters} = 35.8 \, \text{mL} = 0.0358 \, \text{L}
\][/tex]
- Use the ideal gas law [tex]\( PV = nRT \)[/tex] to find the moles of hydrogen gas ([tex]\(n\)[/tex]):
- [tex]\(P = 0.9881 \, \text{atm}\)[/tex],
- [tex]\(V = 0.0358 \, \text{L}\)[/tex],
- [tex]\(R = 0.0821 \, \text{L·atm/(mol·K)}\)[/tex],
- [tex]\(T = 300.15 \, \text{K}\)[/tex].
[tex]\[
n = \frac{PV}{RT} = \frac{0.9881 \times 0.0358}{0.0821 \times 300.15} \approx 0.00144 \, \text{moles of H}_2
\][/tex]
3. Find Moles of Aluminum Using Stoichiometry:
- According to the balanced equation, 3 moles of H₂ come from 2 moles of Al.
- Therefore, moles of Al reacting is:
[tex]\[
\text{Moles of Al} = \left(\frac{2}{3}\right) \times 0.00144 \approx 0.000957 \, \text{moles of Al}
\][/tex]
4. Calculate Mass of Aluminum:
- Molar mass of aluminum = 26.98 g/mol.
- Calculate the mass of aluminum:
[tex]\[
\text{Mass of Al} = 0.000957 \times 26.98 = 0.0258 \, \text{grams}
\][/tex]
Therefore, approximately 0.0258 grams of aluminum react.
1. Understand the Chemical Reaction:
- The reaction between aluminum (Al) and hydrochloric acid (HCl) produces aluminum chloride (AlCl₃) and hydrogen gas (H₂).
- The balanced equation is:
[tex]\[
2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2
\][/tex]
- From the equation, we can see that 2 moles of aluminum produce 3 moles of hydrogen gas.
2. Use the Ideal Gas Law to Find Moles of Hydrogen Gas:
- Given:
- Volume of hydrogen gas = 35.8 mL
- Temperature = [tex]\(27^\circ C\)[/tex]
- Pressure = 751 mmHg
- First, convert the temperature to Kelvin:
[tex]\[
\text{Temperature in Kelvin} = 27 + 273.15 = 300.15 \, \text{K}
\][/tex]
- Convert the pressure from mmHg to atm:
[tex]\[
\text{Pressure in atm} = \frac{751 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.9881 \, \text{atm}
\][/tex]
- Convert the volume from mL to L:
[tex]\[
\text{Volume in liters} = 35.8 \, \text{mL} = 0.0358 \, \text{L}
\][/tex]
- Use the ideal gas law [tex]\( PV = nRT \)[/tex] to find the moles of hydrogen gas ([tex]\(n\)[/tex]):
- [tex]\(P = 0.9881 \, \text{atm}\)[/tex],
- [tex]\(V = 0.0358 \, \text{L}\)[/tex],
- [tex]\(R = 0.0821 \, \text{L·atm/(mol·K)}\)[/tex],
- [tex]\(T = 300.15 \, \text{K}\)[/tex].
[tex]\[
n = \frac{PV}{RT} = \frac{0.9881 \times 0.0358}{0.0821 \times 300.15} \approx 0.00144 \, \text{moles of H}_2
\][/tex]
3. Find Moles of Aluminum Using Stoichiometry:
- According to the balanced equation, 3 moles of H₂ come from 2 moles of Al.
- Therefore, moles of Al reacting is:
[tex]\[
\text{Moles of Al} = \left(\frac{2}{3}\right) \times 0.00144 \approx 0.000957 \, \text{moles of Al}
\][/tex]
4. Calculate Mass of Aluminum:
- Molar mass of aluminum = 26.98 g/mol.
- Calculate the mass of aluminum:
[tex]\[
\text{Mass of Al} = 0.000957 \times 26.98 = 0.0258 \, \text{grams}
\][/tex]
Therefore, approximately 0.0258 grams of aluminum react.
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