9.03 X molecules of butane is present in 9.213 grams of the sample.
1.5 atoms of hydrogen are present in the sample 9.123 grams of butane.
Explanation:
Data given:
mass of butane C4H10= 9.123 grams
atomic mass of butane = 58.12 grams/mole
number of molecules =?
number of hydrogen atoms in the sample=?
number of moles is calculated as :
number of moles =
putting the values in the above equation:
number of moles of butane =
= 0.15 moles
number of molecules = number of moles x 6.02 x (Avagadro number)
Putting the values in above formula:
number of molecules of Butane = 0.15 X 6.023 X
= 9.03 X molecules
Number of hydrogen atoms:
1 mole of C4H10 contains 10 atoms of H
0.15 moles of C4H10 will have x moles
=
x = 0.15 atoms of hydrogen
Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 3 moles of
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of
Thus 0.59 moles of give =
of
Mass of
Thus 31.8 g of will be produced from the given masses of both reactants.
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52.8 grams of fat would be contained in 0.97 lbs of ground beef
