Answer:
Kc = 3.94
Explanation:
CH₃COOH(g) + C₂H₅OH(g) → CH₃COOC₂H₅(g) + H₂O(g)
Liquids aren't included in the equilibrium constant, but at 100°C, all four compounds are a gas. So the equilibrium equation is:
Kc = [CH₃COOC₂H₅] [H₂O] / ([CH₃COOH] [C₂H₅OH])
Set up an ICE table for each row and calculate the value of Kc. Then average the results. (See picture.)
Kc ≈ (3.919 + 4.012 + 3.902) / 3
Kc ≈ 3.94
Answer:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
60 g C2H6 × 1 mol C2H6 x 7 mol O2 x 32 g O2 = ~223.5 g O2
30.068 g 2 mol C2H6 1 mol O2
