Respuesta :
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 = 0,1122 g of C ≡ 112,2mg of C
In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = 0,0110 g of H ≡ 11,0mg of H
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 = [tex]\frac{112,2 mgC}{200,0mg}[/tex]×100 = 56,1%C
%H = [tex]\frac{11,0 mgH}{200,0mg}[/tex]×100 = 5,5%H
%Cl = [tex]\frac{55,14 mgCl}{200,0mg}[/tex]×100 = 27,6%Cl
%N = [tex]\frac{21,66 mgN}{200,0mg}[/tex]×100 = 10,8%N
I hope it helps!
The herbicide has a percent composition of carbon 56 %, hydrogen 27.5 %, Cl 27.7 %, and N 11%.
The moles of gas in 22.4 L of volume is 1 mole. 1 mole of a gas is equivalent to the molar mass of the gas. Thus, the mass of gas at 22.4 L is equivalent to the molar mass.
Computation for the percent composition of herbicide
The volume of carbon dioxide produces is 209.2 mL.
The molar mass of carbon is 12 g.
The mass of carbon dioxide in 209.2 mL is
[tex]\rm 22400\;mL=1\;mol\\209.2\;mL=\dfrac{1}{22400}\;\times\;209.2\;mol\\ 209.2\;mL=0.009\;mol[/tex]
The moles of carbon in a mole of carbon dioxide are the same.
Thus, the mass of carbon is:
[tex]\rm 1\;mol=12\;g\\0.009\;mol=12\;\times\;0.009\;g\\0.009\;g=0.112 g[/tex]
The mass of carbon produced is 0.112 g.
The volume of water vapor produced is 122 mL.
The molar mass of hydrogen is 1 g/mol
The mass of water vapors produced is:
[tex]\rm 22400\;mL=1\;mol\\\122\;mL=\dfrac{1}{22400}\;\times\;122\;mol\\ \\122\;mL=0.005\;mol[/tex]
The moles of water vapor produced is 0.05 mol.
The moles of hydrogen are half the moles of water. Thus, the moles of hydrogen are 0.055 mol.
The mass of hydrogen produced is 0.0025 g.
The mass of Cl in the sample is 55.41 mg.
The total mass of herbicide is 200 mg, or 0.2 g.
The mass of nitrogen in the sample is:
[tex]N=Total -C+H+Cl\\N=0.2-0.112+0.055+0.05514\;g\\N=0.022\;g[/tex]
The mass of nitrogen in the sample is 0.022 g.
The percent composition of the herbicide is given as:
- [tex]\rm C=\dfrac{0.112}{0.2}\;\times\;100\\ C=56\;\%[/tex]
- [tex]\rm H=\dfrac{0.055}{0.2}\;\times\;100\\ H=27.5\;\%[/tex]
- [tex]\rm Cl=\dfrac{55.14}{200}\;\times\;100\\ Cl=27.7\;\%[/tex]
- [tex]\rm N=\dfrac{0.22}{0.2}\;\times\;100\\ N=11\;\%[/tex]
Learn more about percent composition, here:
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