Answer B, because clear ,pure substances makes more sense then others, i searched it all up and came up b, please give brainliest!! good luck
the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
0.00193 mol/L
Given that:
numbers of moles of H₂S = 0.59 moles
Volume = 3.0-L
Equilibrium constant = 9.30 × 10⁻⁸
The equation for the reaction is given as :
2H₂S ⇄ 2H₂(g) + S₂(g)
The initial concentration of H₂S =
The initial concentration of H₂S =
= 0.1966 mol/L
The ICE table is shown be as :
2H₂S ⇄ 2H₂(g) + S₂(g)
Initial 0.9166 0 0
Change -2 x +2 x + x
Equilibrium (0.9166 - 2x) 2x x
(since 2x < 0.1966 if solved through quadratic equation)
The equilibrium concentration for H₂(g) = 2x
∴
= 0.00193 mol/L
Thus, the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
To know more about equilibrium concentration
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Answer:
1.95mol
Explanation:
First let us generate the equation for the reaction:
2H2 + O2 —> 2H2O
From the equation,
1mole of O2 reacted to produce 2moles of H2O.
Therefore, Xmol of O2 will react to produce 3.9 mol of H2O i.e
Xmol of O2 = 3.9/2 = 1.95mol
