How many mL of 0.75M HCl are needed to neutralize 20.0 grams of NaOH?
see below
To determine how many mL of 0.75M HCl are needed to neutralize 20.0 grams of NaOH, we need to use the concept of molar ratios and balanced chemical equations.
1. Write a balanced chemical equation for the neutralization reaction between HCl and NaOH:
HCl + NaOH → NaCl + H2O
2. Calculate the number of moles of NaOH:
Moles = mass / molar mass
The molar mass of NaOH (sodium hydroxide) is 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol.
Moles of NaOH = 20.0 g / 39.99 g/mol ≈ 0.50 mol
3. Determine the molar ratio between NaOH and HCl from the balanced chemical equation:
From the balanced equation: 1 mol NaOH reacts with 1 mol HCl.
4. Therefore, the number of moles of HCl required is also 0.50 mol.
5. To find the volume of 0.75M HCl needed, we can use the equation:
Moles = Molarity × Volume (in liters)
Rearrange the equation to solve for volume:
Volume = Moles / Molarity
Plugging in the known values:
Volume = 0.50 mol / 0.75 mol/L
= 0.67 L
6. Convert the volume from liters to milliliters:
1 L = 1000 mL
Volume (in mL) = 0.67 L × 1000 mL/L
= 670 mL
Therefore, approximately 670 mL of 0.75M HCl is needed to neutralize 20.0 grams of NaOH.