Comments (4) on “How many hydrogen atoms are contained in 167 grams of propane?”
Propane is C3H8
So Molar mass propane = 3*12 + 8*1 = 44 No. of moles of propane = 167/44 = 3.795 moles
From the stoichiometry: No. of moles of hydrogen = 3.795 moles * (8 moles hydrogen / 1 mole propane) = 30.36 moles
Using Avogadros number, 1 mole of any substance contains 6.022 x 10^23 atoms, therefore: No. of hydrogen atoms = 30.36 * 6.022 x 10^23 = 1.828 x 10^25 atoms
Propane is C3H8
So Molar mass propane = 3*12 + 8*1 = 44
No. of moles of propane = 167/44 = 3.795 moles
From the stoichiometry:
No. of moles of hydrogen = 3.795 moles * (8 moles hydrogen / 1 mole propane) = 30.36 moles
Using Avogadros number, 1 mole of any substance contains 6.022 x 10^23 atoms, therefore:
No. of hydrogen atoms = 30.36 * 6.022 x 10^23 = 1.828 x 10^25 atoms
po-215 i got it right on the quiz
hloride, what is the mass of calcium in
explanation:
9.03 X [tex]10^{22}[/tex] 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 = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
putting the values in the above equation:
number of moles of butane = [tex]\frac{9.213}{58.12}[/tex]
= 0.15 moles
number of molecules = number of moles x 6.02 x [tex]10^{23}[/tex] (Avagadro number)
Putting the values in above formula:
number of molecules of Butane = 0.15 X 6.023 X [tex]10^{23}[/tex]
= 9.03 X [tex]10^{22}[/tex] molecules
Number of hydrogen atoms:
1 mole of C4H10 contains 10 atoms of H
0.15 moles of C4H10 will have x moles
[tex]\frac{10}{1}[/tex] = [tex]\frac{x}{0.15}[/tex]
x = 0.15 atoms of hydrogen