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In a reaction of 40.0 grams of sodium hydroxide with 60.0 grams of sulfuric acid. How...

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persianimmortal | Student, Grade 12 | (Level 1) Valedictorian

Posted April 17, 2013 at 10:14 PM via web

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In a reaction of 40.0 grams of sodium hydroxide with 60.0 grams of sulfuric acid. How much sodium sulfate is produced?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted April 17, 2013 at 11:14 PM (Answer #1)

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The balanced chemical equation for this reaction can be written as:

`2 NaOH + H_2SO_4 -> Na_2SO_4 + 2 H_2O`

We need to look for the limiting reagent first before we can be able to determine the amount of sodium sulfate produced. 

`40.0 grams NaOH * (1 mol e NaOH)/(39.997 grams) * (1 mol e Na_2SO_4)/(2 mol es NaOH)`

= 0.500 moles `Na_2SO_4`

 

`60.0 grams H_2SO_4 * (1 mol e H_2SO_4)/(98.079 grams) * (1 mol e Na_2SO_4)/(1 mol e H_2SO_4)`

= 0.612 moles `Na_2SO_4`

 

Looking at the results, the limiting reagent is NaOH and therefore, we will use the value derived from the amount of NaOH. 

`0.500 mol es Na_2SO_4 * (142.04 grams Na_2SO_4)/(1 mol e Na_2SO_4)`

= 71.0 grams `Na_2SO_4` -> answer

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