Question

How many hydrogen atoms are in 7.80 mol of ammonium sulfide?

Answer 1

The number  of hydrogen atoms that are in 7.80 moles of ammonium sulfide are  3.756 x 10^25  atoms

calculation

Step  1:  calculate  the  moles of H  in  (NH4)2S

 since there are 8 atoms of  H in (NH4)2S  the moles of

   H =  8 x7.80 =  62.4  mole

 Step  2: use of Avogadro's law  constant  to  determine the number  of  H atom

          that is 1 mole = 6.02 x10^23 atoms

                     62.4 moles=?  atoms

• cross  multiplication

= [ 62.4  x 6.02 x10^23]/  1 mole= 3.756  x10^25    atoms

Answer 2

Answer: There are $3.75 \times 10^{25}$ number of H atoms in 7.80 mol of ammonium sulfide.

Explanation:

The formula of ammonium sulfide is $(NH_{4})_{2}S$.

There are 2 nitrogen atoms, 8 hydrogen atoms and 1 sulfur atom in one molecule of ammonium sulfide.

Calculate moles of hydrogen atoms as follows.

              $8 \times 7.80$ = 62.4

According to Avogadro's number 1 mole contains $6.022 \times 10^{23} atoms$

Therefore, calculate number of H atoms as follows.

               $62.4 \times 6.022 \times 10^{23}$

             = $375.77 \times 10^{23}$

             = $3.75 \times 10^{25}$

Thus, there are $3.75 \times 10^{25}$ number of H atoms in 7.80 mol of ammonium sulfide.