Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent composition of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
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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