How many moles of ammonium ions are in 6.965 g of ammonium carbonate? Express your answer using four significant figures.

Answer:0.1450

there are 3.63*10⁻² moles of ammonium anions

Explanation:

since the ammonium carbonate is

(NH₄) ₂ (CO₂)

and the ammonium carbonate dissociate according to

(NH₄) ₂ (CO₂) →2*NH₄⁺ + CO₂²⁻ (we will neglect other carbonate species for simplicity)

then each mole of ammonium carbonate generates 2 moles of ammonium anions, then

n AC = 2 * n A

where n represents moles AC represents ammonium carbonate and A represents ammonium anions

also we know that

n AC = m AC / M AC

where

m = mass of ammonium carbonate = 6.595 g

M = molecular weight of ammonium carbonate = 96.09 g/mol

then

n AC = m AC / M AC = 6.595 g / 96.09 g/mol = 7.25*10⁻² moles of ammonium carbonate

thus

n AC = 2 * n A → n A = n AC / 2 = 7.25*10⁻² / 2 = 3.63*10⁻² moles of ammonium anions