How many moles of ${(N H_{4})}_{2} S O_{4}$ $(NH_4)_2SO_4$ are present in 25.0 grams of ${(N H_{4})}_{2} S O_{4}$ $(NH_4)_2SO_4$ ?

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Answer 1 · 2016-04-13T05:50:40

0.2 moles

Mass of one mole of Ammonium Sulfate is 132 g

We can set up conversion factors/ ratios

1 mol Ammonium Sulfate = 132g of $(N$ $(N$ H4 $)_{2}$ $)_2$ S $O_{4}$ $O_4$

1 mol Ammonium Sulfate / 132g of $(N$ $(N$ H4 $)_{2}$ $)_2$ S $O_{4}$ $O_4$ (a)

and

132g of $(N$ $(N$ H4 $)_{2}$ $)_2$ S $O_{4}$ $O_4$ / 1 mol of Ammonium Sulfate (b)

Using the conversion factor (a) we can find number of moles.

25 g $(N$ $(N$ H4 $)_{2}$ $)_2$ S $O_{4}$ $O_4$ x 1 mol of Ammonium sulfate / 132g of $(N$ $(N$ H4 $)_{2}$ $)_2$ S $O_{4}$ $O_4$

0.2 moles Ammonium sulfate

How many moles of (NH4)2SO4(NH_4)_2SO_4 are present in 25.0 grams of (NH4)2SO4(NH_4)_2SO_4?