Calculate the mass of 4.95 x 10^15 Ag atoms

              8.87 × 10⁻⁷ grams

Explanation:

Moles = No. of Particles / 6.022 × 10²³ Particles/mol

Moles = 4.95 × 10¹⁵ atoms / 6.022 × 10²³ atoms/mol

Moles = 8.22 × 10⁻⁹ mol

Also,

Mass = Moles × M.Mass

Mass = 8.22 × 10⁻⁹ mol × 107.87 g/mol

Mass = 8.87 × 10⁻⁷ grams

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

a:

by the combination of an unlimited number of elements