The chemical reaction is
C3H8O + (9/2) O2 = 3CO2 + 4H2O
Theoretical amount of oxygen needed = 0.1 mol of propanol ( 4.5 mol O2 / 1 mol propanol) = 0.45 mol
The amount of oxygen used in mol can be calculated using the ideal gas equation
PV = nRT
@room conditions
P = 1 atm
T = 25C ~ 298 K
R = 0.08205 mol – L/ atm – K
1 dm3 = 1 L
(1 atm) (12 L) = n (0.08205)*(298K)
n = 0.49 mol O2
Excess O2 = 0.49 – 0.45 = 0.04 mol O2
Amount of CO2 produced = 0.1 mol propanol (3 mol CO2/ 1 mol propanol) = 0.3 mol
Amount of H2) produced = 0.1 mol propanol (4 mol H2O/ 1 mol propanol) = 0.4 mol
TOTAL amount of gases = 0.04 + 0.3 + 0.4 = 0.74 mol
Therefore
V = (0.74*0.08205*298)/1
V = 18.0 dm3, the answer is letter D
Answer:The the correct answer is option (A).
Explanation:
12 L of oxygen gas burns 0.10 mol of propanol.(1 L = 1)
According to reaction, 2 moles of propanol gives 6 moles of then, 0.10 moles of propanol will give: moles of that is 0.30 mol.
The final volume of the gas evolved after the reaction can be determined by Ideal gas equation. Since, the reaction is taking place at room temperature the value of pressure and temperature will be:
At room temperature:
Pressure = 1 atm
Temperature , T= 293 K
The closest answer from the given options is option (A) that is 7.20
Answer:
1. balanced
2.unbalanced
3.unbalabced
4.balanced
2HCl yields H2 + Cl2
2K + 2HCl yields 2KCl + H2
SO3 + H2O yields H2SO4
Answer: Based on experimental data
Explanation: The hypothesis is not the ultimate proven theory, the more a preliminary assumption on the basis of which further research is carried out with the help of experiments in order to prove the hypothesis as accurate. If the experiments carried out showed that the hypothesis is wrong, it is necessary to revise the hypothesis and set up a new experiment.
Ah yes.
The atomic radius is basically the distance between the nucleus of the atom and the outermost electron in the farthest orbital of the atom. The more bigger the atomic radius , the less powerful is the force that holds together the nucleus and its electrons.