Answer:
%C = 56,1%
%H = 5,5%
%Cl = 27,6%
%N = 10,8%
Explanation:
The moles of CO₂ are the same than moles of C in the herbicide.
Moles of H₂O are ¹/₂ of moles of H in the herbicide.
Moles of CO₂ are obtained using:
n = PV/RT
Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K
moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>
In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>
As you have 55,14 mg of Cl, the mg of N are:
200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N
Thus, precent composition of the herbicide is:
%C = ×100 = 56,1%C
%H = ×100 = 5,5%H
%Cl = ×100 = 27,6%Cl
%N = ×100 = 10,8%N
I hope it helps!