If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require _____ mL of 0.2389 M NaOH for neutralization.
A) 31.22
B) 39.98
C) 7.80
D) 79.96
E) 15.61

Please give an answer and an explanation, thanks!

To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.

Moles H₂SO₄ (Molar mass: 98.08g/mol):

0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =

3.7289x10⁻³moles H₂SO₄

And moles of NaOH that you require to neutralize the acid are:

3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =

7.4578x10⁻³ moles NaOH

Using a 0.2389M NaOH solution:

7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL

Right answer is:

If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require _____ mL of 0.2389 M NaOH for neutralization. A) 31.22 B) 39.98 C) 7.80 D) 79.96 E) 15.61 Please give an answer and an explanation, thanks!

Respuesta :

A) 31.22

Otras preguntas

If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require _____ mL of 0.2389 M NaOH for neutralization.
A) 31.22
B) 39.98
C) 7.80
D) 79.96
E) 15.61

Please give an answer and an explanation, thanks!