The moles of disodium edta used is calculated using the below formula
moles =molarity x volume in liters
molarity=0.050m
volume in liters = 22/1000=0.022 L
moles is therefore= 0.022 x0.050 =1.1 x10^-3 moles of disodium edta
Answer:
V₂ = 19.3 L
Explanation:
Given data:
Initial volume = 29.3 L
Initial pressure = standard = 1 atm
Initial temperature = standard = 273 K
Final temperature = -23°C (-23+273 = 250K)
Final volume = ?
Final pressure = 1.39 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1 atm × 29.3 L × 250 K / 273 K × 1.39 atm
V₂ = 7325 atm .L. K / 379.47 K.atm
V₂ = 19.3 L
To determine the amount of excess reactant left, we need to first identify which is the limiting and the excess reactant. We do as follows:
<span>A + 2B → AB2
2 mol A ( 2 mol B / 1 mol A) = 4 mol B needed to be consumed completely
5 mol B ( 1 mol A / 2 mol B) = 2.5 mol A needed to be consumed completely
Therefore, the limiting would be reactant A and the excess would be reactant B. The amount of the excess would be:
5 mol B - 4 mol B = 1 mol B <----------LAST OPTION</span>
Answer: -
4C₂H₅ + 13 O₂ → 8 CO₂ + 10 H₂O
Explanation: -
The given equation is
C₂H₅ + O₂ → CO₂ + H₂O
Balancing for H,
2C₂H₅ + O₂ → CO₂ + 5H₂O
Balancing for C
2C₂H₅ + O₂ → 4CO₂ + 5H₂O
Now right hand side we are getting 13 O which is an odd number.
In order to balance O, we first try to even out the number of O on right side by multiplying whole equation by 2.
4C₂H₅ + 2O₂ → 8CO₂ + 10H₂O
Now balancing for O
4C₂H₅ + 13 O₂ → 8 CO₂ + 10 H₂O