C3H8 + O2 (please give me brainliest$
Answer:
A) t = 22.5 min and B) t = 29.94 min
Explanation:
Initial concentration, [A]₀ = 100
Final concentration = 100 -75 = 25
Time = 45 min
A) First order reaction
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[25] − ln[100] = - 45k
-1.386 = -45k
k = 0.0308 min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[50] − ln[100] = - 0.0308 * t
-0.693 = -0.0308 * t
t = 22.5 min
B) Zero Order
[A] = [A]₀ − kt
Using the values from the initial reaction and solving for k, we have;
25 = 100 - k(45)
-75 = -45k
k = 1.67 M min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
[A] = [A]₀ − kt
50 = 100 - (1.67)t
-50 = - 1.67t
t = 29.94 min
Answer:
58.9g of SO2 is produced
8g of oxygen remains unconsumed
Explanation:
The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:
CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)
Molar mass of CS2 = 76.139 g/mol
Molar mass of O2 = 15.99 g/mol
Molar mass of SO2 = 64.066 g/mol
Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles
Number of moles of O2 = 30g/15.999 g/mol =1.88 moles
From the chemical reaction
1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2
Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2
Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced
thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2
Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining
Or 8g of oxygen
58.9g of SO2 is produced
oxygen is the limiting
Here you go! Feel free to ask questions!
