How many mL of 0.75M HCl are needed to neutralize 20.0 grams of NaOH?
moles NaOH = grams/molar mass
moles NaOH = moles HCl (1:1 in the reaction but write the equation to verify.)
M = mols/L. You have M and molesk solve for L, then convert to mL.
To solve this question, we need to determine the number of moles of NaOH and the stoichiometry between HCl and NaOH.
First, let's determine the number of moles of NaOH:
1. Find the molar mass of NaOH:
- The atomic mass of Na (sodium) is approximately 23 grams/mol.
- The atomic mass of O (oxygen) is approximately 16 grams/mol.
- The atomic mass of H (hydrogen) is approximately 1 gram/mol.
Therefore, the molar mass of NaOH is:
(1 x 23) + (1 x 16) + (1 x 1) = 39 grams/mol.
2. Convert the given mass of NaOH to moles:
- Divide the given mass of NaOH (20.0 grams) by the molar mass of NaOH (39 grams/mol).
- 20.0 grams ÷ 39 grams/mol ≈ 0.513 moles of NaOH.
Now, let's determine the stoichiometry between HCl and NaOH:
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
From the equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of water and one mole of sodium chloride.
Finally, let's calculate the volume (in mL) of 0.75M HCl needed to neutralize 0.513 moles of NaOH:
1. Use the mole-to-mole ratio from the balanced equation to convert moles of NaOH to moles of HCl:
- Since the stoichiometry is 1:1, the moles of HCl required will also be 0.513 moles.
2. Calculate the volume of 0.75M HCl using the molarity formula:
- Molarity (M) = moles/volume (L)
- Rearrange the formula to solve for volume:
Volume (L) = moles/Molarity
- Substitute the values into the formula:
Volume (L) = 0.513 moles / 0.75 M ≈ 0.684 L
3. Convert volume from liters to milliliters:
- Multiply the volume in liters by 1000:
0.684 L x 1000 mL/L ≈ 684 mL
Therefore, approximately 684 mL of 0.75M HCl are needed to neutralize 20.0 grams of NaOH.