There is 1 mol N/1 mol NaNO3; therefore there must be .....mole N in 2.6 mol of NaNO3.
I may be incorrect here - as this contradicts with the previous answer, but this is how I would tackle this problem. One mole of nitrogen has a mass of 14 g/mole (which can be seen on the periodic table). In one mole of NaNO3 (sodium nitrate), the mass is 85g/mole (adding all the masses of each atom together: shown below)
Mass of Na = 23
Mass of N = 14
Mass of O3 = 16 (3) = 48
23 + 14 + 48 = 85
Since there are 2.6 moles of NaNO3, we multiply 85 by 2.6.
85 * 2.6 = 221
Then, we divide by 14, as that is the mass of one nitrogen atom, and we want to see the number of moles of nitrogen there are in 2.6 moles of NaNO3.
221/14 = 15.7857.......
221/14 = 15.79
So, there are around 15.79 moles of nitrogen in 2.6 moles of NaNO3.
X is correct to a point.
There are 85*2.6 = 221 g NaNO3. BUT there are NOT 14 g N in that. %N in NaNO3 is (14/85)*100 = about 16.47% so in 221 g NaNO3 there are 221*0.1647 = about 36.40 g N or 36.40/14 = 2.60 mols N.
Of course that's the long way to go. You can see 1 N atom/1 molecule NaNO3; therefore, 1 mole N/1 mole NaNO3 and 2.6 mols N in 2.6 mols NaNO3.