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How many grams of ammonium sulfate can be produced if 2500.0 g of H2SO4 react with excess NH3 according to the equation 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)? Calculate the percent yield if 3100 g (NH4)2SO4 were actually produced.
-Calculate the theoretical yield of ammonium sulfate.
-Calculate the percent yield for the associated question
-Write the unit for the theoretical yield (see related question)

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Oseni

If 2500 grams of H2SO4 react with excess NH3 according to the illustration, the amount of ammonium sulfate that can be produced would be 3,370.92 grams. If only 3100 grams of ammonium sulfate were produced, the percentage yield would be 91.96%.

Stoichiometric calculations

From the equation of the reaction:

2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)

The mole ratio of H2SO4 and (NH4)2SO4 is 1:1

Mole of 2500 grams H2SO4 = 2500/98 = 25.51 moles

Equivalent mole of (NH4)2SO4 = 25.51 moles

Mass of 25.51 moles (NH4)2SO4 = 25.51 x 132.14 = 3,370.92 grams

If 3100 g of (NH4)2SO4 were actually produced, the percentage yield would be:

Percent yield = 3100/3370.92 x 100 = 91.96%

More on percent yield can be found here: https://brainly.com/question/17042787

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