What is the reduction half equation of Fe(s)+ 2 HC2H3O2(aq) → Fe(C2H3O2)2(aq) + H2(g)
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Answer:
The total pressure in the flask is 0,619 atm.
Explanation:
For the reaction:
2CH₄(g) + 3Cl₂(g) ⟶ 2 CH₃Cl(g) + 2HCl(g) + 2Cl⁻(g)
The moles of CH₄ in 295 mL at STP are:
n = PV/RT
Where P is pressure (1 atm), V is volume (0,295L), R is gas constant (0,082atmL/molK) and T is temperature (273,15 K)
Replacing, moles of CH₄ are <em>0,0132 moles</em>
In the same way, moles of chlorine are <em>0,0324 moles</em>
As 3 moles of Cl₂ react with 2 moles of CH₄, for a total reaction of 0,0132 moles of CH₄ you need:
0,0132 moles CH₄ × = <em>0,0198 moles Cl₂. </em>That means that 0,0324-0,0198 = <em>0,0126 moles of Cl₂ are in excess.</em>
As the reaction reaches in 77%, the moles of CH₄ that don't react are:
0,0132×(100%-77%)= <em>3,036x10⁻³ moles of CH₄</em>
Also, the moles of Cl₂ that don't react are:
0,0126 + 0,0198×(100%-77%)= <em>0,0172 moles of Cl₂</em>
The moles produced of each compound are:
0,0132×77% × = <em>0,0102 moles of CH₃Cl -</em><em>that are the same moles of HCl and Cl⁻</em><em>-</em>
Thus, total moles in the flask are:
<em>3,036x10⁻³ moles of CH₄ + 0,0172 moles of Cl₂ + 0,0102 moles of CH₃Cl + 0,0102 moles of HCl + 0,0102 moles of Cl⁻ = </em><em>0,0507 total moles</em>
As the volume of the flask is 2,00L and the final temperature is 298 K. The total pressure in the flask is:
P = nRT/V
<em>P = 0,619 atm</em>
<em> </em>
I hope it helps!