Calculate the mass percent of sulfuric acid in a solid sample given a 0.8390g sample of that solid required 32.22 mL of 0.2012 M NaOH for neutralization.
A) 62.11
B) 2.64
C) 37.89
D) 75.78
E) 1.27

Please give an answer and an explanation, thanks!

Answers

Answer 1

Answer

C

Explanation:

i remember learning this

Answer 2

Mass percentage does not have any unit as numerator and denominator contains the same unit. The percentage of sulfuric acid in solid sample is 38.07%.

What is percentage by mass?

Mass percentage represents the the percentage of each element that is making a particular compound.

Mathematically,

Percentage of mass = (component’s mass ÷ total mass) x 100%

[tex]\rm 2 NaOH aq + H _{2} SO _{4} aq → Na _{2}SO _{4}aq + 2 H _{2} O[/tex]

Volume of NaOH solution = 32.22 mL = 0.03222 L

Concentration or molarity of the NaOH solution = 0.2012 M

Moles of NaOH : 0.03222×0.2012=0.006444mol

2 moles of sodium hydroxide reacts with 1 mole of sulfuric acid.

0.006444mol of sodium hydroxide reacts with =0.006444÷2=0.003222mol of sulfuric acid.

Mass of sulfuric acid neutralized = 0.003222 moles × 98.08 g/mol =0.31601376g

Percentage of sulfuric acid in solid sample= (0.31601376g÷0.8390g)×100=38.07%

Thus the percentage of sulfuric acid in solid sample is 38.07%.

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