Question

Given 4.80g of ammonium carbonate, find:
Number of moles of the compound
Number of moles of ammonium ions
Number of moles of carbonate ions
Number of moles of hydrogen atoms
Number of hydrogen atoms

Answer 1

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

1) Number of moles of the compound:

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

2) Number of moles of ammonium ions :

• Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ = (2.0)(0.05 mol) =  0.1 mol.

3) Number of moles of carbonate ions :

• Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

4) Number of moles of hydrogen atoms:

• Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO₃ = (8.0)(0.05 mol) = 0.4 mol.

5) Number of hydrogen atoms:

• It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

Using cross multiplication:

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

∴ The no. of atoms in  0.4 mol of H atoms = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = 2.4 x 10²³ molecules.

Answer 2

The number of moles of the compound  =0.05 mol

The number of moles of ammonium ions =0.1 mol

The number of moles of carbonate ions =0.05 mol

The number of moles of hydrogen atoms =0.4 mol

The number of hydrogen atoms =$\bold{ 2.4 \times 10^2^3\ molecules}$

What is ammonium carbonate?

Ammonium chloride is a solid crystalline salt with a chemical formula   $\bold{(NH_4)_2CO_3}$. It slowly decomposes and produces ammonia.

Given, Mass is 4.80 g

a. The number of moles of the compound:

$\bold{Number\; of\; moles = \dfrac{mass}{molar mass} = \dfrac{4.80 g}{96.09 g/mo} = 0.05 mol}$

b. Number of moles of ammonium ions :

 Ammonium carbonate is decomposing according to this reaction

$\bold{(NH_4)_2CO_3 = 2NH_4^+ + CO_32^-}$

The number of moles of NH₄⁺ ions in 0.05 mol of $\bold{(NH_4)_2CO_3}$ is

$\bold{(2.0)\times(0.05 mol) = 0.1 mol}$

c. Number of moles of carbonate ions :

The number of moles of CO₃²⁻ ions in 0.05 mol of $\bold{(NH_4)_2CO_3}$  is

$\bold{(1.0)\times(0.05 mol) = 0.05 mol}$

d. the number of moles of hydrogen atoms:

The number of moles of H atoms in 0.05 mol of $\bold{(NH_4)_2CO_3}$  is

$\bold{(8.0)\times(0.05 mol) = 0.4 mol}$

e. Number of hydrogen atoms:

Every mole of a molecule or element contains Avogadro's number ($6.022 \times 10^2^3$) of molecules or atoms.

By multiplication:

1.0 mole of H atoms contains $6.022 \times 10^2^3$ atoms.

0.4 mole of H atoms contains how many atoms.

The number of atoms in 0.4 mol of H atoms

$\bold{ (6.022 \times 10^2^3)\dfrac{(0.4 mole)}{(1.0 mole)} = 2.4 \times 10^2^3\ molecules}$

Thus, a. -0.05 mol, b. -0.1 mol, c. -0.05 mol, d. -0.4 mol, e. -$\bold{ 2.4 \times 10^2^3\ molecules}$.

Learn more about ammonium carbonate, here:

https://brainly.com/question/2022471