The antacid Amphogel contains aluminum hydroxide Al(OH)3. How many milliliters of 6.00M HCl are required to react with 150mL of 1.12M Al(OH)3? Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(aq)
Help please and thanks in advance
This is an example of a simple stoichiometry problem (as opposed to a limiting reagent problem) and all of these are worked alike.
Step 1. Write and balance the equation. You have that.
Step 2. Convert mols Al(OH)3 to mols. mol = M x L. (Note: in cases where grams are given, then mol = grams/molar mass.)
Step 3. Using the coefficients in the balanced equation, convert mols Al(OH)3 to mols HCl.
Step 4. Now convert to L. M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.
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To find out how many milliliters of 6.00M HCl are required to react with 150mL of 1.12M Al(OH)3, we need to use the concept of stoichiometry. The balanced equation tells us that 1 mole of Al(OH)3 reacts with 3 moles of HCl.
First, let's calculate the moles of Al(OH)3 using the given volume and concentration:
moles of Al(OH)3 = volume (in liters) × concentration
= 150 mL × (1 L / 1000 mL) × 1.12 mol/L
= 0.168 moles
Since the molar ratio between Al(OH)3 and HCl is 1:3, we know that 0.168 moles of Al(OH)3 will react with 0.168 moles × (3 moles HCl / 1 mole Al(OH)3) = 0.504 moles of HCl.
Now we can determine the volume of 6.00M HCl needed to react with 0.504 moles of HCl:
volume of HCl = moles of HCl / concentration
= 0.504 mol / 6.00 mol/L
≈ 0.084 L
Finally, convert the volume from liters to milliliters by multiplying by 1000:
volume of HCl = 0.084 L × (1000 mL / 1 L)
= 84 mL
Therefore, approximately 84 milliliters of 6.00M HCl are required to react with 150mL of 1.12M Al(OH)3.