how many milliliters of .2 M sodium hydroxide are needed to completely neutralize 25.0g of sulfuric acid

To determine the amount of sodium hydroxide (NaOH) needed to neutralize sulfuric acid (H2SO4), you need to balance the chemical equation between these two compounds.

The reaction equation between sodium hydroxide and sulfuric acid is as follows:

2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

From the balanced equation, you can observe that one mole of sulfuric acid reacts with two moles of sodium hydroxide. This means that the stoichiometric ratio between sodium hydroxide and sulfuric acid is 2:1.

Now, let's calculate the number of moles of sulfuric acid present in 25.0g.

To do this, you need to divide the given mass of sulfuric acid by its molar mass. The molar mass of sulfuric acid is calculated as follows:

H2SO4 = 2(1.01 g/mol) + 32.06 g/mol + 4(16.00 g/mol) = 98.09 g/mol

Moles of sulfuric acid = Mass / Molar mass
Moles of sulfuric acid = 25.0g / 98.09 g/mol

Next, we can use the stoichiometric ratio to determine how many moles of sodium hydroxide are needed to neutralize the calculated moles of sulfuric acid.

Since the stoichiometric ratio is 2:1 (NaOH to H2SO4), we can say that the moles of NaOH required will be half the moles of H2SO4.

Moles of NaOH = Moles of H2SO4 / 2

Finally, to determine the volume of the sodium hydroxide solution needed, we need to use the concentration of sodium hydroxide.

The concentration of sodium hydroxide is given as 0.2 M, which means that there are 0.2 moles of NaOH in every liter of solution.

Volume of NaOH solution (in liters) = Moles of NaOH / Concentration (in M)

Since we want the answer in milliliters (mL), we can convert liters to milliliters by multiplying the volume in liters by 1000.

Volume of NaOH solution (in mL) = Volume of NaOH solution (in liters) x 1000

By following these calculations, you can determine the number of milliliters of 0.2 M sodium hydroxide needed to completely neutralize 25.0g of sulfuric acid.