Hello!
We know that by the Law of Avogrado, for each mole of substance we have 6.02 * 10²³ atoms, if:
The molar mass of water (H2O)
H = 2 * (1u) = 2u
O = 1 * (16u) = 16u
---------------------------
The molar mass of H2O = 2 + 16 = 18 g / mol
If:
1 mol we have 6.02 * 10²³ atoms
1 mole of H2O we have 18 g
Then we have:
18 g ------------- 6.02 * 10²³ atoms
5 g -------------- x
I Hope this helps, greetings ... DexteR! =)
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol