Fermentation
got it right on the test :)
Answer:
n₂ = 2.55 mol
Explanation:
Given data:
Initial number of moles = 0.758 mol
Initial volume = 80.6 L
Final volume = 270.9 L
Final number of moles = ?
Solution:
Formula:
V₁/n₁ = V₂/n₂
V₁ = Initial volume
n₁ = initial number of moles
V₂ = Final volume
n₂ = Final number of moles
now we will put the values in formula.
V₁/n₁ = V₂/n₂
80.6 L / 0.758 mol = 270.9 L/ n₂
n₂ = 270.9 L× 0.758 mol / 80.6 L
n₂ = 205.34 L.mol /80.6 L
n₂ = 2.55 mol
Molar mass :
Li₂S = <span>45.947 g/mol
AlCl</span>₃ = <span>133.34 g/mol
</span><span>3 Li</span>₂<span>S + 2 AlCl</span>₃<span> = 6 LiCl + Al</span>₂S₃
3 * 45.947 g Li₂S ----------> 2 * <span>133.34 g AlCl</span>₃
1.084 g Li₂S ----------------> ?
Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947
Mass Li₂S = 289.08112 / 137.841
Mass Li₂S = 2.0972 g
hope this helps!