Answer:a) 11.34 g of ethane can be formed
b) is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1.
2.
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.378 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
moles of left = (2.10-0.378) = 1.72 moles
mass of left=
According to stoichiometry :
As 1 mole of give = 1 mole of
Thus 0.378 moles of give = of
Mass of
Thus 11.34 g of ethane is formed.
Answer:
a. 75 mol O2
Explanation:
1 mol C2H6O -> 3 mol O2
25 mol C2H6O -> x
x= (25 mol C2H6O * 3 mol O2)/ 1 mol C2H6O
x= 75 mol O2
Answer : The concentration of AB after 14.0 s is, 0.29 M
Explanation :
The expression used for second order kinetics is:
where,
k = rate constant =
t = time = 14.0 s
= final concentration = ?
= initial concentration = 1.50 M
Now put all the given values in the above expression, we get:
Therefore, the concentration of AB after 14.0 s is, 0.29 M
True. They form neap tides. neap tides are lower than their regular and can get to the lowest point.