Question

How many ml of 1.498 M h2so4 is needed to neutralize 45.8 g of naoh

Answer 1

To neutralize 45.8 g of NaOH with 1.498 M H2SO4, approximately 383 mL of H2SO4 is needed.

Step-by-step explanation:

To find the volume of 1.498 M H2SO4 needed to neutralize 45.8 g of NaOH, we can use the balanced chemical equation:

H2SO4 (aq) + 2NaOH(aq) → Na2SO4 (aq) + 2H2O (l)

We can calculate the number of moles of NaOH:

1. 45.8 g NaOH × (1 mol NaOH / 40.00 g NaOH) = 1.145 mol NaOH

Since the stoichiometric ratio between H2SO4 and NaOH is 1:2, we need twice the number of moles of H2SO4:

1. 1.145 mol NaOH × (1 mol H2SO4 / 2 mol NaOH) = 0.573 mol H2SO4

Finally, we can calculate the volume of 1.498 M H2SO4:

1. 0.573 mol H2SO4 × (1 L H2SO4 / 1.498 mol H2SO4) = 0.383 L = 383 mL