Answer: Metals form cations.
The alkali metals (the IA elements) lose a single electron to form a cation with a 1+ charge.
The alkaline earth metals (IIA elements) lose two electrons to form a 2+ cation.
Aluminum, a member of the IIIA family, loses three electrons to form a 3+ cation.
Therefore, metals in the s and p block of the periodic table have 1, 2 or 3 electrons in their outermost orbit (or valence shell). Now to gain a stable octet metals lose either 1, 2 or 3 electrons from the valence shell thus forming cation with +1, +2 or +3 charge.
Answer:
3.00 mol
Explanation:
Given data:
Mass of P₄ = 211 g
Mass of oxygen = 240 g
Moles of P₂O₅ = ?
Solution:
Chemical equation:
P₄ + 5O₂ → 2P₂O₅
Number of moles of P₄:
Number of moles = mass/ molar mass
Number of moles = 211 g / 123.88 g/mol
Number of moles = 1.7 mol
Number of moles of O₂ :
Number of moles = mass/ molar mass
Number of moles = 240 g / 32g/mol
Number of moles = 7.5 mol
Now we will compare the moles of product with reactant.
O₂ : P₂O₅
5 : 2
7.5 : 2/5×7.5 = 3.00
P₄ : P₂O₅
1 : 2
1.7 : 2×1.7 = 3.4 mol
Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.
Mass of P₂O₅:
Mass = number of moles × molar mass
Mass = 3 mol ×283.9 g/mol
Mass = 852 g
Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.
Explanation: Reaction to form alum from Aluminium is given as:
We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.
By stoichiometry,
2 moles of Al is producing 2 moles of Alum
Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol
Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol
54 g/mol of aluminium will produce 948 g/mol of alum, so
Amount of Alum produced = 17.34 grams
Theoretical yield of alum = 17.34 grams.
