(PA 32) I\'ve repeatedly gotten this question wrong. I\'ve tried rounding to all

Question

(PA 32) I've repeatedly gotten this question wrong. I've tried rounding to all kinds of sig figs including not rounding at all. Help please!

You want to neutralize 21.8 mol of sulfuric acid with 6.0 M sodium hydroxide. How much 6.0 M sodium hydroxide must you add to react exactly with the sulfuric acid?

Selected Answer: 260L

Response Feedback:

Did you write out the balanced equation?

Did you use the mole ratio to convert from mol sulfuric acid to mol sodium hydroxide?

Did you multiply mol sodium hydroxide by 1 L/6.0 mol (1/M)?

(PA 32) I've repeatedly gotten this question wrong. I've tried rounding to all kinds of sig figs including not rounding at all. Help please!

You want to neutralize 21.8 mol of sulfuric acid with 6.0 M sodium hydroxide. How much 6.0 M sodium hydroxide must you add to react exactly with the sulfuric acid?

Selected Answer: 260L

Response Feedback:

Did you write out the balanced equation?

Did you use the mole ratio to convert from mol sulfuric acid to mol sodium hydroxide?

Did you multiply mol sodium hydroxide by 1 L/6.0 mol (1/M)?

Explanation / Answer

balanced equation:

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

1 mol             2 mol

1 mol H2SO4 ---------------------> 2 mol NaOH

21.8 mol H2SO4 -----------------> x mol NaOH

x = 2 x 21.8 /1

x = 43.6

moles of H2SO4 = 43.6

molarity of NaOH = 6 M (given)

molarity = moles / volume

6 = 43.6 / volume

volume = 43.6 / 6

volume = 7.27 L

volume of NaOH needed = 7.27 L

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