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
Answer:
m= 29.645 g
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
density of wood = 0.77 g/cm³
volume= 38.5 cm³
mass= ?
Solution:
d= m/v
m= d × v
m= 0.77 g/cm³× 38.5 cm³
m= 29.645 g
The solution to the problem is as follows:
<span>a)
C + O2 = CO2
Molar mass CO2 = 44 g/mol
3.67 g CO2 * 1 mol / 44 g =
=0.0834 mol CO2 = 0.0834 mol C
I hope my answer has come to your help. God bless and have a nice day ahead!
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Answer:A chlorine atom in its ground state has a total of seven electrons in orbitals related to the atoms third energy level.
Explanation: