How many grams of NaOH needed to completely neutralize 3L of 1.75M HCL

209.98 g of NaOH

Explanation:

We are given;

Volume of HCl as 3 L Molarity of HCl as 1.75 M

We are required to calculate the mass of NaOH required to completely neutralize the acid given.

First, we write a balanced equation for the reaction between NaOH and HCl

That is;

NaOH + HCl → NaCl + H₂O

Second, we determine the number of moles of HCl

Number of moles = Molarity * Volume

= 1.75 M * 3 L

= 5.25 moles

Third, we use the mole ratio to determine the moles of NaOH

From the reaction,

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = Moles of HCl

= 5.25 moles

Fourth, we determine the mass of NaOH

Molar mass of NaOH = 39.997 g/mol

Mass of NaOH = 5.25 moles * 39.997 g/mol

= 209.98 g

Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl