The oxidation state of potassium ion K = +1
The oxidation state of oxygen ion O = -2
So, the oxidation state of O2 is = -2 x 2 = -4
Since, KBrO2 is neutral so,
(+1) + (x) + (-4) = Zero
-3 + X = Zero
So, X = +3
The oxidation state of individual bromine atom in KBrO2 is +3
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol
18g C*1mol C/12 g C = 1.5 mol C
C + O2 → CO2
from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol
1.5 mol O2*32 g O2/1 mol O2 = 48 g O2
In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is </span><span>2.) 54 g.</span>
Answer:
ΔH = -976.5 kJ
Explanation:
For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.
The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).
So, 24.3 g of benzene has :
n = mass/ molar mass
n = 24.3/78.11
n = 0.311 moles
2 moles ------------ -6278 kJ
0.311 moles ----------- x
By a simple direct three rule:
2x = -1953.08
x = -976.5 kJ
