How many moles of ammonium ions are in 0.705 g of ammonium carbonate?

Answer 1

Ammonium chloride is a soluble salt.For every mole of ammonium carbonate, 2 moles of ammonium ions and 1 mole of trioxocarbonate(iv) ions are produced after dissolution.
0.705/96=0.00734 moles.Since the mole ratio between ammonium chloride and ammonium ions is 1:2,therefore the number of moles of ammonium ions is 0.01468 moles.

Answer 2

The formula for ammonium carbonate is (NH4)2 CO3. The molecular weight is 2x18+60, or 96. The number of moles of ammonium carbonate present is 0.705/96 (you do the math -- my calculator needs batteries). Each mole provides two moles of ammonium ion, so double what you just got and you're done.

Answer 3

ammonium carbonate -( NH4)2CO3

number of moles of ammonium carbonate = 0.705g/(14+14+8+12+16+16+16) = 0.00734 (3 sig. fig)

ammonium carbonate dissociates to form ammonium ions and carbonate ions.

for every molecule of ammonium carbonate that dissociates, you get 2 mol of ammonium ions and 1 mol of carbonate ions.

thus, for 0.00734 mol of ammonium carbonate, you get

0.00734*2=0.0147 mol ( 3 sig. fig) of ammonium ions.

Answer 4

ammonium carbonate = (NH4)2CO3
molar mass of (NH4)2CO3 = 96.11 g/mol
mass of (NH4)2CO3 = 0.705 g
(0.705 g)/(96.11 g/mol) = 0.00734 mol (NH4)2CO3

each mol of (NH4)2CO3 has 2 mol NH4
2 * 0.00734 mol = 0.0147 mol NH4