How many mole are in 2.84*10^22 molecules of h2
1 answer:
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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.