Answer:
0.00915 M of 
remain after 5.16 seconds.
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
s⁻¹
Initial concentration = 0.054 M
Final concentration = ? M
Time = 5.16 s
Applying in the above equation, we get that:-
![[A_t]=0.054e^{-0.344\times 5.16}\ M=\frac{1\times \:0.054}{e^{1.77504}}\ M=0.00915\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.054e%5E%7B-0.344%5Ctimes%205.16%7D%5C%20M%3D%5Cfrac%7B1%5Ctimes%20%5C%3A0.054%7D%7Be%5E%7B1.77504%7D%7D%5C%20M%3D0.00915%5C%20M)
<u>0.00915 M of remain after 5.16 seconds.</u>
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g
Answer: -2.13%
Explanation:
I will assume that the correct time is 94 seconds, since it is stated as a fact. The measured time of 92 seconds is -2 seonds from the actual. That gives a percent error of:
[(Measured-Actual)/Actual] or (-2/94) or -2.13%
