<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
Answer:
-973 KJ
Explanation:
The balanced reaction equation is;
N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)
Reduction potential of hydrazine = -1.16 V
Reduction potential of chlorine = 1.36 V
From;
E°cell= E°cathode - E°anode
E°cell= 1.36 - (-1.16)
E°cell= 2.52 V
∆G°=- nFE°cell
n= number of moles of electrons = 4
F= Faraday's constant = 96500 C
E°cell = 2.52 V
∆G°=- (4 × 96500 × 2.52)
∆G°= -972720 J
∆G°= -972.72 KJ
Answer:
An addition reaction
Step-by-step explanation:
In an addition reaction, two or more molecules come together to form a single product, for example,
C₂H₂ + 2Cl₂ ⟶ C₂H₂Cl₄
This reaction consists of two successive additions. The product of the first reaction becomes a reactant and adds a second molecule of Cl₂ to form C₂H₂Cl₄
C₂H₂ + Cl₂ ⟶ <em>C₂H₂Cl₂
</em>
<em><u>C₂H₂Cl₂</u></em><u> + Cl₂ ⟶ C₂H₂Cl₄
</u>
C₂H₂ + 2Cl₂ ⟶ C₂H₂Cl₄
Answer:

Explanation:
To solve this problem, we can use the Combined Gas Laws:

Data:
p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C
p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 1 + 273.15) K = 274.15 K
T₂ = (82 + 273.15) K = 355.15 K
(b) Calculate the new pressure
