Consider the equation: 4NH3+7O2---> 4NO2 + 6 H2O
a. How many grams of oxygen are necessary to produce 4.50 moles of
NO2?
b. How many molecules of water are produced when 2.25 moles of ammonia
are completely reacted?

Hi teatea!
a. From the given equation, it's clear that you get 4 NO2 molecules
for every 7 O2 molecules reacted. Thus,

7 moles of O2 -----> 4 moles of NO2
X moles of O2 -----> 4.5 moles of NO2

Now, solving for X, we get that X=7*4.5/4= 7.875. We need 7.875 moles
of O2 to produce 7 moles of NO2. Finally, we need to find how many
grams of O2 are 7.875 moles of O2. In order to do this, we muust first
calculate the molecular mass of O2. The atomic mass of oxigen (O) is
16 (we know this from the periodic table). Therefore, the molecular
mass of O2 is 16*2=32 grams per mole. Knowing this number, we can
easily calculate how many grams of O2 we need:

7.875 * 32 = 252 grams of O2.

b. The procedure is eaxactly the same as in point (a). For every 4
molecules of ammonia (NH3), we get 6 molecules of water (H2O). Thus,

4 moles of NH3 -----> 6 moles of H2O
2.25 moles of NH3 --> X moles of H2O

Then, solving for X, we get X=(2.25/4)*6=3.375 moles of H2O. So the
answer is that we get 3.375 moles of water when we completely react
2.25 moles of molecules of NH3.


If you have any further question regarding this topic, please don't
hesitate to request a clarification.

Best wishes!
elmarto