Answer:
A. 2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)
B. 761.42 L
Explanation:
A. Step 1:
The equation for the reaction.
C12H26(l) + O2(g) —> CO2(g) + H2O(g)
A. Step 2:
Balancing the equation.
The equation can be balance as follow:
C12H26(l) + O2(g) —> CO2(g) + H2O(g)
There are 12 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 12 in front of CO2 as illustrated below:
C12H26(l) + O2(g) —> 12CO2(g) + H2O(g)
There are 26 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 13 in front of H2O as illustrated below:
C12H26(l) + O2(g) —> 12CO2(g) + 13H2O(g)
Now, there are a total of 37 atoms of O2 on the right side and 2 atoms on the left. It can be balance by putting 37/2 in front of O2 as illustrated below:
C12H26(l) + 37/2O2(g) —> 12CO2(g) + 13H2O(g)
Multiply through by 2 to clear the fraction from the equation.
2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)
Now the equation is balanced
B. Step 1:
We'll by obtaining the number of mole of C12H26 in 0.450 kg of C12H26. This is illustrated below:
Molar Mass of C12H26 = (12x12) + (26x1) = 144 + 26 = 170g/mol
Mass of C12H26 = 0.450 kg = 0.450x1000 = 450g
Number of mole of C12H26 =?
Number of mole = Mass/Molar Mass
Number of mole of C12H26 = 450/170
Number of mole of C12H26 = 2.65 moles
B. Step 2:
Determination of the number of mole of CO2 produced by the reaction. This is illustrated below:
2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)
From the balanced equation above,
2 moles of C12H26 produced 24 moles of CO2.
Therefore, 2.65 moles of C12H26 will produce = (2.65x24)/2 = 31.8 moles of CO2.
B. Step 3:
Determination of the volume of CO2 produced by the reaction.
Pressure (P) = 1 atm
Temperature (T) = 19°C = 19°C + 273 = 292K
Gas constant (R) = 0.082atm.L/Kmol
Number of mole (n) = 31.8 moles
Volume (V) =?
The volume of CO2 produced by the reaction can b obtained by applying the ideal gas equation as follow:
PV = nRT
1 x V = 31.8 x 0.082 x 292
V = 761.42 L
Therefore, the volume of CO2 produced is 761.42 L
Answer:
a. [tex]C_{12}H_{26}(l)+\frac{37}{2}O_2(g)\rightarrow 12CO_2(g)+13H_2O(g)[/tex]
b. [tex]V=761 L[/tex]
Explanation:
Hello,
a. In this case, the required balanced combustion is:
[tex]C_{12}H_{26}(l)+\frac{37}{2}O_2(g)\rightarrow 12CO_2(g)+13H_2O(g)[/tex]
b. In this case, we compute the available moles of dodecane as:
[tex]n_{C_{12}H_{26}}^{available}=0.450kgC_{12}H_{26}*\frac{1000gC_{12}H_{26}}{1kgC_{12}H_{26}}*\frac{1molC_{12}H_{26}}{170gC_{12}H_{26}} =2.65molC_{12}H_{26}[/tex]
Next, we compute the yielded moles of carbon dioxide by using the 1:12 mole ratio given at the chemical reaction:
[tex]n_{CO_2}=2.65molC_{12}H_{26}*\frac{12molCO_2}{1molC_{12}H_{26}} =31.8molCO_2[/tex]
Finally, by using the ideal gas equation we compute the volume at the given conditions:
[tex]V=\frac{nRT}{p}=\frac{31.8mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{1atm}\\\\V=761 L[/tex]
Best regards.
