Respuesta :

Answer:

23.92 g

Explanation:

Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

I mole of H2SO4 = 2 moles of NaOH

24.5/82 = 24.5/82 × 2

= 0.598 moles of NaOH will neutralize

Mass= mole× molar mass

Molar mass of NaOH= 23+16+1 = 40g/mol

Mass= 0.598 × 40 = 23.92g of NaOH

Answer:

20g

Explanation:

First of all, you have got your formula wrong. It should be [tex]H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O[/tex].

Start by finding the Relative Formula Mass for sulfuric acid. [tex]H_{2}SO_{4}[/tex] is equal to (2x1)+32+(16x4)=98. You then calculate Mols by doing [tex]Mols=\frac{Mass}{Relative Formula Mass}[/tex]. So, [tex]\frac{24.5}{98} = 0.25[/tex], so the Mols of Sulfuric Acid is 0.25.

The Mol ratio of Sulfuric Acid to Sodium Hydroxide is 1:2, so 0.25:0.5.

Now, you have to find the Relative Formula Mass of Sodium Hydroxide, or [tex]NaOH[/tex]. 23+16+1=40. Finally, multiply 0.5 by 40, and you'll find that the maximum mass is 20g.