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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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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