FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
Explanation:
The given data is as follows.
Pressure (P) = 760 torr = 1 atm
Volume (V) = = 0.720 L
Temperature (T) = = (25 + 273) K = 298 K
Using ideal gas equation, we will calculate the number of moles as follows.
PV = nRT
Total atoms present (n) =
=
= 0.0294 mol
Let us assume that there are x mol of Ar and y mol of Xe.
Hence, total number of moles will be as follows.
x + y = 0.0294
Also, 40x + 131y = 2.966
x = 0.0097 mol
y = (0.0294 - 0.0097)
= 0.0197 mol
Therefore, mole fraction will be calculated as follows.
Mol fraction of Xe =
=
= 0.67
Therefore, the mole fraction of Xe is 0.67.
Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+
⇄
Formation constant Kf
Kf = / ( [
][
] ) = 5.0 × 10¹⁰
Now,
[] =
; ∝₄ = 0.35
so the equilibrium is;
+
⇄
+ 4H⁺
Given that; =
{ 1 mol
reacts with 1 mol
}
so at equilibrium, =
= x
∴
+
⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [
][
] ) =
/ ( [
][
] ) = 5.0 × 10¹⁰
⇒ / ( [
][
] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷