Calculate the mass of water having the same number of hydrogen atoms as are present in 32 gm of methane.

2 answers

1 mol CH4 has 6.02E23 molecules
1 mol H2O has 6.02E23 molecules
so 1 mol CH4 has same number molecules as 1 mol H2O.

mols methane = g/molar mass = 32/16 = 2 mol CH4. That has 2*4 = 8 mols H atoms.
mols H2O vs CH4; You must multiply mols H in H2O by 2 to get mols H in CH4. So 2 mol H2O x 2 = 4 mols H2O and 4*18 = 72 g H2O. Chech that.

32g CH4 is 32/16 = 2 mols CH4 which is 2*4 = 8 mols H atoms.
72g H2O is 72/18 = 4 mols H2O which is 4*2 = 8 mols H atoms.
Bingo!
Ans:- 72 gram
Solution:-

here,
At first for methane(CH4):

N.O of mole = given mass(32)/molar mass(16) [molar mass = (12+4), given
mass= 32]
n.o of moles = 2 moles
we know that,
If 1 mole of CH4molecule = 4 moles of H atom. then,
2 moles of CH4 molecules = 8 moles of H atom
Again,
1 mole of H ====> 6.022*10^23 atoms of H. Then,
8 mole of H ====> 6.022*10^23 * 8====>4.8176*10^24 atoms of H.

for mass for H:-
6.022*10^23 atoms of H ====>1 gm of H then
4.8176*10^24 atoms of H====> 8 gm of H.

For mass of O:-
Here to get the number of oxygen(O2) in water(H2O), we have to half the number of atom of the atoms that we get from the H. It is because, look in the molecular formula of H2O there are 2 atoms of hydrogen(H) and one atom of oxygen(O). so, me must understand that the one atom of oxygen uses two atom of hydrogen(H).

therefore,
no of oxygen atom= (4.8176*10^24/ 2)= 2.4088*10^24 atoms

finally, For oxygen(O):-
if 6.022*10^23 atoms ====> 16 gram of O then
2.4088*10^24 atoms =====> (16/6.022*10^23)*2.4088*10^24====>64- gram

hence, the total mass or weight of water = 64+ 8 = 72 gram.
I hope you have understood! Thank You!