<span>Molecular compounds, which are represented by molecules, are usually made of non-metals only (or of metalloids and non-metals). Ionic compounds, which are represented by formula units, are made of metals and non-metals.
More detail if you're interested: Molecules and formula units are the representative particles for molecular and ionic compounds, respectively. By that I mean, one unit of a molecular compound is a molecule...a bundle of atoms covalently bonded that exists separately from all the other molecules. One unit of an ionic compound is a formula unit. A formula unit is a representation of the compound's formula. For example, the formula unit of NaCl is one Na^+1 ion and one Cl^-1 ion. The formula unit of AlCl3 is one Al^+3 ion and three Cl^-1 ions. Ionic compounds don't have separate bundles of atoms like molecular compounds do, so the formula unit is just the smallest number of ions that it takes to represent the formula. </span>
Answer:
is the concentration of 
in the solution.
Explanation:
![Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq%29%20%2B%203%20en%5Crightleftharpoons%20%5BNi%28en%29_3%5D%5E%7B2%2B%7D%28aq%29)
Concentration of nickel ion = ![[Ni^{2+}]=x](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D%5D%3Dx)
Concentration of nickel complex= ![[[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L](https://tex.z-dn.net/?f=%5B%5BNi%28en%29_3%5D%5E%7B2%2B%7D%5D%3D%5Cfrac%7B0.16%20mol%7D%7B2%20L%7D%3D0.08%20mol%2FL)
Concentration of ethylenediamine = ![[en]=\frac{0.80 mol}{2 L}=0.40 mol/L](https://tex.z-dn.net/?f=%5Ben%5D%3D%5Cfrac%7B0.80%20mol%7D%7B2%20L%7D%3D0.40%20mol%2FL)
The formation constant of the complex = 
The expression of formation constant is given as:
![K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%5B%5BNi%28en%29_3%5D%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D%5D%5Ben%5D%5E3%7D)
is the concentration of in the solution.
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
The formula for Hexafluoride is F6S.
