Answer:
a) 0.525 mol
b) 0.525 mol
c) 0.236 mol
Explanation:
The combustion reactions (partial and total) will be:
C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O
C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O
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2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O
It means that the reaction will form 50% of each gas.
a) 0.525 mol of CO
b) 0.525 mol of CO₂
c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol
So, the number of moles is the mass divided by the molar mass:
n = 11.5/100 = 0.115 mol
For the stoichiometry:
2 mol of C₇H₁₆ -------------- (37/2) mol of O₂
0.115 mol of C₇H₁₆ --------- x
By a simple direct three rule:
2x = 2.1275
x = 1.064 mol of O₂
Which is the moles of oxygen that reacts, so are leftover:
1.3 - 1.064 = 0.236 mol of O₂
Since you did not give a calculation,
I will just give an example. Suppose that you are to burn 5 kg of methane (CH4)
from 0 to 10°C. The specific heat capacity of methane is 4.475 kJ/kg-K.
H
= mCpT
H
= (5kg)( 4.475 kJ/kg-K)(10-0)
H = 223.75 kJ
Because the enthalpy is positive
in value, methane takes in heat.
The answer is ( The concentrations of the products and reactants do not change.)
B, water and ice are the main causes .
The oxidation half equation is Zn ------> Zn^2+ + 2e while the reduction half equation is Cu^2+ + 2e------> Cu.
A redox reaction is a reaction in which there is a loss/gain of electrons. The specie that gives out electrons experiences an increase in oxidation number while the specie that gains the electrons experiences a decrease in oxidation number.
For the reaction; CuCl2 + Zn → ZnCl2 + Cu
The oxidation half equation is;
Zn ------> Zn^2+ + 2e
The reduction half equation is;
Cu^2+ + 2e------> Cu
The chloride ion is excluded because its oxidation number does not change from left to right in the reaction.
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