Question #e25c7

Ammonium sulfide, #(NH_4)_2S#, is an ionic compound formed when two ammonium cations, #NH_4^(+)#, form an ionic bond with one sulfide anion, #S^(2-)#.

This means that every mole of ammonium sulfide will contain 2 moles of ammonium cations and 1 mole of sulfide ions.

Since you have 4.560 moles of ammonium sulfide, you'll also have

#4.560cancel("moles"(NH_4)_2S) * ("2 moles "NH_4^(+))/(1cancel("mole"(NH_4)_2S)) = color(green)("9.120 moles "NH_4^(+))#

To determine how many grams of ammonium sulfide would contain this many moles, use the ion's molar mass

#9.120cancel("moles") * "18.04 g"/(1cancel("mole")) = color(green)("164.5 g "NH_4^(+))#

ALTERNATIVE APPROACH

You can also determine how many grams of ammonium ions you'd get by using percent composition.

#(2 * 18.04cancel("g/mol"))/(68.154cancel("g/mol")) * 100 = "52.94%"# #NH_4^(+)#

This means that 100 g of ammonium sulfide will contain 52.94 g of ammonium ions.

The mass of ammonium sulfide would be

#4.560cancel("moles") * "68.154 g"/(1cancel("mole")) = "310.78 g"#

This means that you get

#310.78cancel("g"(NH_4)_2S) * ("52.94 g "NH_4^(+))/(100cancel("g"(NH_4)_2S)) = color(green)("164.5 g "NH_4^(+))#