How many grams of hydrogen are in "69" g of ch4o?

             8.62 g of Hydrogen

Solution:

Molar mass of CH₄O (Methanol) is 32 g.mol⁻¹.

It means,

                  32 g of CH₄O contains  =  4 g of Hydrogen

Then,

             69 g of CH₄O will contain  =  X g of Hydrogen

Solving for X,

                    X  =  (69 g × 4 g) ÷ 32 g

                    X  =  8.62 g of Hydrogen

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.

what are the two models

explanation: