Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:
![C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl](https://tex.z-dn.net/?f=C_2H_6%2BCl_2%5Crightarrow%20C_2H_5Cl%2BHCl)
Moles of ethane = ![\frac{300.0 g}{30 g/mol}=10 mol](https://tex.z-dn.net/?f=%5Cfrac%7B300.0%20g%7D%7B30%20g%2Fmol%7D%3D10%20mol)
Moles of chlorine gases =![\frac{650.0 g}{71 .0 g/mol}=9.1549 mol](https://tex.z-dn.net/?f=%5Cfrac%7B650.0%20g%7D%7B71%20.0%20g%2Fmol%7D%3D9.1549%20mol)
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g
![\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5C%25%20Yield%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
![\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%](https://tex.z-dn.net/?f=%5C%25Yield%20%28C_2H_5Cl%29%3D%5Cfrac%7B490.0%20g%7D%7B590.4910%20g%7D%5Ctimes%20100%3D82.98%5C%25)
The percent yield of chloro-ethane in the reaction is 82.98%.
Answer: 5.84x10^4
Explanation: You start with (9.00x10^26 atoms)x(1 mol/ 6.022x10^23)x(39.10 g/1 mol)
Answer:
1
Explanation:
single displacement replaces one element with another and it's product is a pure element.
Isotopes are atoms of different elements which have the same mass.