Answer:
1) 0.05 mol.
2) 0.1 mol.
3) 0.05 mol.
4) 0.4 mol.
5) 2.4 x 10²³ molecules.
Explanation:
<em>1) Number of moles of the compound:</em>
no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.
<em>2) Number of moles of ammonium ions
:</em>
- Ammonium carbonate is dissociated according to the balanced equation:
<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>
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.
<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) = <em>0.1 mol.</em>
<em>3) Number of moles of carbonate ions
:</em>
- Ammonium carbonate is dissociated according to the balanced equation:
<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>
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.
<em>4) Number of moles of hydrogen atoms:</em>
- 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.
<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>
<em>5) Number of hydrogen atoms:</em>
- It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.
<u><em>Using cross multiplication:</em></u>
1.0 mole of H atoms contains → 6.022 x 10²³ atoms.
0.4 mole of H atoms contains → ??? atoms.
<em>∴ The no. of atoms in 0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>
= 0.78 mol Mg₃(Si₂O₅)₂(OH)₂