Question

PLS HELPPP How many moles of oxygen atoms (not molecules) are present in 6.41×1025 molecules of dinitrogen pentoxide (N2O5)? 1. 532 mol 2. 106 mol 3. 32 × 1025 mol 4. 266 mol

Answer 1

Answer:

(4) 266 moles

Explanation:

We have Dinitrogen Pentoxide N2O5

6.41*10^25 molecules are given

No of Moles of N2O5 = $\frac{No of Molecules }{Avogadro's Number}$

                                     = $\frac{6.41 * 10^25}{6.022*10^23}$

                                      = 106.5 mol

Now using Unitary Method

2 Mole of Nitrogen pentoxide require 5 mole of Oxygen to form N2O5

1 mole of N = $\frac{5}{2}*O$

In 106.5 mole of N  = $\frac{5}{2}*106.5$ = 266.25 mole

So, 6.41*10^25 molecules of N2O5 will require 266.25 mole of Oxygen atoms.