How many mL of 1.2 M HCl is needed to react with 5.8 g of Al(OH)3
Technically any number you choose. I'm sure the author of the problem meant (but didn't say) how much HCl is need to react COMPLETELY with the 5.8 g Al(OH)3
3HCl + Al(OH)3 ==> AlCl3 + 3H2O
mols Al(OH)3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al(OH)3 to mols HCl. That's
?mols Al(OH)3 x (3 mols HCl/1 mol Al(OH)3) = ? mols Al(OH)3 x 3/1 = ?
Then M = mols/L.
You know M of the HCl, and mols HCl, solve for L HCl, then convert to mL.
Well, let's do the math and find out!
First, we need to convert the mass of Al(OH)3 into moles. The molar mass of Al(OH)3 is 78 g/mol. So, 5.8 g / 78 g/mol = 0.0744 mol.
The balanced equation for the reaction between HCl and Al(OH)3 is:
3HCl + Al(OH)3 -> AlCl3 + 3H2O
Looking at the equation, we can see that it takes 3 moles of HCl to react with 1 mole of Al(OH)3.
So, if we have 0.0744 mol of Al(OH)3, we would need 3 * 0.0744 mol = 0.2232 mol of HCl.
To calculate the volume in mL, we need to use the molarity of HCl. The molarity (M) of HCl is given as 1.2 M, which means 1.2 moles of HCl is present in 1 liter of solution.
Since we have 0.2232 moles of HCl, we can calculate the volume in liters using the equation:
Volume (L) = moles / molarity
Volume (L) = 0.2232 mol / 1.2 M = 0.186 L
Finally, to convert liters to milliliters, simply multiply the volume by 1000.
Volume (mL) = 0.186 L * 1000 mL/L = 186 mL
Therefore, you would need 186 mL of 1.2 M HCl to react with 5.8 g of Al(OH)3.
To determine the volume of 1.2 M HCl needed to react with 5.8 g of Al(OH)3, we'll follow these steps:
Step 1: Calculate the molar mass of Al(OH)3.
Step 2: Convert the mass of Al(OH)3 to moles.
Step 3: Use the balanced chemical equation to determine the stoichiometry between Al(OH)3 and HCl.
Step 4: Use the stoichiometry to find the number of moles of HCl needed.
Step 5: Calculate the volume of 1.2 M HCl needed using the molarity-moles-volume relationship.
Let's go step by step.
Step 1: Calculate the molar mass of Al(OH)3.
Al(OH)3 has aluminum (Al) with a molar mass of 26.98 g/mol, and three hydroxide (OH) groups, each with an oxygen (O) with a molar mass of 16.00 g/mol, and a hydrogen (H) with a molar mass of 1.01 g/mol.
So the molar mass of Al(OH)3 is:
(1 x Al) + (3 x OH)
= (1 x 26.98 g/mol) + (3 x (16.00 g/mol + 1.01 g/mol))
= 27.98 g/mol + 3 x 17.01 g/mol
= 27.98 g/mol + 51.03 g/mol
= 79.01 g/mol
Step 2: Convert the mass of Al(OH)3 to moles.
Given:
Mass of Al(OH)3 = 5.8 g
Molar mass of Al(OH)3 = 79.01 g/mol
Number of moles of Al(OH)3 = Mass / Molar mass
= 5.8 g / 79.01 g/mol
= 0.0734 mol
Step 3: Use the balanced chemical equation to determine the stoichiometry between Al(OH)3 and HCl.
The balanced chemical equation for the reaction between Al(OH)3 and HCl is:
Al(OH)3 + 3HCl -> AlCl3 + 3H2O
From the equation, it is clear that 1 mole of Al(OH)3 reacts with 3 moles of HCl.
Step 4: Use the stoichiometry to find the number of moles of HCl needed.
Number of moles of HCl = Number of moles of Al(OH)3 x (3 moles HCl / 1 mole Al(OH)3)
= 0.0734 mol x (3 mol HCl / 1 mol Al(OH)3)
= 0.2202 mol HCl
Step 5: Calculate the volume of 1.2 M HCl needed using the molarity-moles-volume relationship.
With a concentration of 1.2 M HCl, we know that there are 1.2 moles of HCl present in 1 liter of solution (1000 mL).
Therefore, the number of moles of HCl needed can be converted to volume using the equation:
Volume (mL) = (Number of moles HCl / HCl concentration) x 1000 mL
Volume (mL) = (0.2202 mol / 1.2 M) x 1000 mL
= 183.5 mL (rounded to the nearest tenth)
Therefore, approximately 183.5 mL of 1.2 M HCl is needed to react with 5.8 g of Al(OH)3.
To determine the volume of 1.2 M HCl needed to react with 5.8 g of Al(OH)3, we will follow these steps:
1. Write and balance the chemical equation for the reaction between HCl and Al(OH)3:
```
Al(OH)3 + 3HCl -> AlCl3 + 3H2O
```
2. Determine the molar mass of Al(OH)3:
- The molar mass of Al is 26.98 g/mol.
- The molar mass of H is 1.01 g/mol.
- The molar mass of O is 16.00 g/mol.
- The molar mass of Al(OH)3 is therefore:
```
(26.98 g/mol × 1) + (1.01 g/mol × 3) + (16.00 g/mol × 3) = 78.00 g/mol
```
3. Convert the given mass of Al(OH)3 to moles:
- Using the molar mass of Al(OH)3 calculated in step 2:
```
Moles of Al(OH)3 = mass of Al(OH)3 / molar mass of Al(OH)3
= 5.8 g / 78.00 g/mol
= 0.0744 mol
```
4. Determine the moles of HCl required:
- According to the balanced equation, 1 mol of Al(OH)3 reacts with 3 mol of HCl.
- Therefore, moles of HCl = 0.0744 mol × 3 = 0.2232 mol
5. Calculate the volume of 1.2 M HCl required:
- The concentration of HCl is given as 1.2 M, which means 1.2 moles of HCl are present in 1 liter of solution.
- Therefore, volume (in liters) = moles of HCl / concentration of HCl
= 0.2232 mol / 1.2 mol/L
= 0.186 L
6. Convert the volume from liters to milliliters:
- 1 liter is equal to 1000 milliliters.
- Therefore, volume (in milliliters) = 0.186 L × 1000 mL/L
= 186 mL
Thus, 186 mL of 1.2 M HCl is needed to react with 5.8 g of Al(OH)3.