What volume (in mL) of 0.300 M HCI would be required to completely react with 3.65 g of Al in the following chemical reaction? 2 Al(s) + 6 HCI(aq) → 2 AICI, (aq) + 3 H, (g)
What volume (in mL) of 0.300 M HCI would be required to completely react with 3.65 g of Al in the following chemical reaction? 2 Al(s) + 6 HCI(aq) → 2 AICI, (aq) + 3 H, (g)
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![### Determining the Volume of 0.300 M HCl Required to React with 3.65 g of Aluminum
**Question:**
What volume (in mL) of 0.300 M HCl would be required to completely react with 3.65 g of Al in the following chemical reaction?
\[
2 \text{Al}(s) + 6 \text{HCl}(aq) → 2 \text{AlCl}_3(aq) + 3 \text{H}_2(g)
\]
**Explanation:**
In this problem, we need to determine how much of a 0.300 M hydrochloric acid (HCl) solution is needed to completely react with 3.65 grams of aluminum (Al).
**Step-by-Step Solution:**
1. **Determine the number of moles of Al:**
The molar mass of Al (aluminum) is approximately 26.98 g/mol.
\[
\text{Moles of Al} = \frac{3.65 \text{ g}}{26.98 \text{ g/mol}} \approx 0.135 \text{ moles}
\]
2. **Use the balanced chemical equation:**
According to the balanced chemical equation:
\[
2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2
\]
This tells us that 2 moles of Al react with 6 moles of HCl. Therefore, 1 mole of Al requires 3 moles of HCl.
\[
\text{Moles of HCl required} = 0.135 \text{ moles Al} \times 3 \frac{\text{moles HCl}}{\text{mole Al}} = 0.405 \text{ moles HCl}
\]
3. **Convert moles of HCl to volume:**
We know the molarity (M) of the HCl solution is 0.300 M, which means 0.300 moles of HCl are in 1 liter (1000 mL) of solution.
\[
\text{Volume (L)} = \frac{\text{moles}}{\text{molarity}} = \frac{0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ae9165-8c64-40e9-9a68-65bc31ca98b7%2F98fe5045-20c3-4a8a-880d-2d716f17bdbb%2F054jhyd_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining the Volume of 0.300 M HCl Required to React with 3.65 g of Aluminum
**Question:**
What volume (in mL) of 0.300 M HCl would be required to completely react with 3.65 g of Al in the following chemical reaction?
\[
2 \text{Al}(s) + 6 \text{HCl}(aq) → 2 \text{AlCl}_3(aq) + 3 \text{H}_2(g)
\]
**Explanation:**
In this problem, we need to determine how much of a 0.300 M hydrochloric acid (HCl) solution is needed to completely react with 3.65 grams of aluminum (Al).
**Step-by-Step Solution:**
1. **Determine the number of moles of Al:**
The molar mass of Al (aluminum) is approximately 26.98 g/mol.
\[
\text{Moles of Al} = \frac{3.65 \text{ g}}{26.98 \text{ g/mol}} \approx 0.135 \text{ moles}
\]
2. **Use the balanced chemical equation:**
According to the balanced chemical equation:
\[
2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2
\]
This tells us that 2 moles of Al react with 6 moles of HCl. Therefore, 1 mole of Al requires 3 moles of HCl.
\[
\text{Moles of HCl required} = 0.135 \text{ moles Al} \times 3 \frac{\text{moles HCl}}{\text{mole Al}} = 0.405 \text{ moles HCl}
\]
3. **Convert moles of HCl to volume:**
We know the molarity (M) of the HCl solution is 0.300 M, which means 0.300 moles of HCl are in 1 liter (1000 mL) of solution.
\[
\text{Volume (L)} = \frac{\text{moles}}{\text{molarity}} = \frac{0.
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