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:
The value of equilibrium constant is 29.45.
Explanation:
Moles of hydrogen gas = 2.00 mol
Concentration of hydrogen gas =
Moles of iodine gas = 1.00 mol
Concentration of iodine gas =
initially
2.00 M 1.00 M 1.00 M
At equilibrium:
(2.00-x/2) (1.00-x/2) x
Moles of HI at equilibrium = 1.80 M
Concentration of HI at equilibrium =
The expression of an equilibrium constant is given by ;
Putting x equal to 1.80 M.
The value of equilibrium constant is 29.45.
Answer:
the Answer is B because that what I know