Answer:
see explanation
Explanation:
To determine limiting reactant divide mole quantities of reactants by the respective coefficient in the balanced equation. The smaller value is the limiting reactant.
P₄ + 5O₂ => 2P₂O₅
12/1 = 12 15/5 = 3
O₂ is the limiting reactant. P₄ will be in excess when rxn stops.
Because the formation of rust is a kind of chemistry reaction.
After chemistry reaction, compound with new properties is produced.
Of course, color is one of the new properties.
For example, if you put iron in colorless acid solution, a green solution is made.
This kind of chemisty reaction differs from physical reaction, for example, pigment with one color dissolves in liquid, a liquid with same color is made.
Answer:

Explanation:
We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.
M_r: 58.32
Mg(OH)₂ + … ⟶ … + 2HOH
m/g: 58.3
(a) Moles of Mg(OH)₂

(b) Moles of H₂O
The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

The reaction will form
of water.
Answer:
the anserw should be 665KJ