Chemistry Help! 4NH(3)+5O(2)---4NO+6H(2)0 Help with Chemical Reactions?

Question

A. How many oxygen atoms are there in 15.00 grams of NO? Answer in Moles
B. What mass of oxygen is needed to react with 8.517 grams of ammonia?
C.How many moles of water are produced in a reaction in which 45.015 grams of NO are produced?
D.How many grams of NO are actually produced when 60.02 grams of ammonia are allowed to react with 96.00 grams of oxygen?

I am so lost and confused with these questions, any help would be greatly appreciated. Thank you!!! :)

Answer 1

A:
The atomic mass of nitrogen (N) is 14.01 (g/mol)
The atomic mass of oxygen (O) is 16.00 (g/mol)
Then the molecular mass of NO is 14.01 + 16.00 = 30.01 (g/mol).
In 15(g) of NO there are 15(g) / 30.01 (g/mol) = 0.500 (mol) molecules.
Since each molecule of NO has 1 oxygen atom, there are 0.500 (mol) of oxygen atoms in 15.00(g) of NO.

B.
The reaction is:
4 NH3 + 5 O2 = 4 NO + 6 H2O
The atomic mass of Hydrogen (H) is: 1.01 (g/mol)
The atomic mass of nitrogen (N) is 14.01 (g/mol)
Then the molecular mass of NH3 is 14.01 + 3 x 1.01 = 17.04 (g/mol).
You have 8.517 (g) of NH3. This is 8.517(g) / 17.04 (g/mol) = 0.500 (mol) of NH3.
From the reaction formula you know that for every 4 moles of NH3 you need 5 moles of O2. This means you need 0.500 (mol) * 5 / 4 = 0.625 (mol) of O2.
The atomic mass of oxygen (O) is 16.00 (g/mol)
The molecular mass of O2 is 2 x 16.00 = 32.00 (g/mol). Therefore you will need 0.625 (mol) * 32.00 (g/mol) = 20.00 (g) of O2.

C)
from A) The molecular mass of NO is 30.01(g/mol)
so 45.015(g) is 45.015(g) / 30.01 (g/mol) = 1.500 (mol) of NO.
From the reaction equation you know that for every 4 moles of NO produced there will be 6 moles of water produced, so there will be: 1.500(mol) * 6 / 4 = 2.250 (mol) of water produced.

D.
From B: Then the molecular mass of NH3 is 17.04 (g/mol). So there are 60.02 (g) / 17.04 (g/mol) = 3.522 (mol) of NH3.
From B: The molecular mass of O2 is 32.00 (g/mol).
So there are 96.00 (g) / 32.00 (g/mol) = 3.000 (mol) of O2

Now comes the tricky part. You are going to have an excess of one of the components and we have to determine which component (NH3 or O2) is the limiting reagent. From the equation you know that for every 4 moles of NH3 you need 5 moles of O2. If NH3 is limiting then there would be at least
3.522 (mol) *5 / 4 = 4.403 (mol) of O2 necessary. Unfortunately you only have 3.000 (mol) of O2. So oxygen is the limiting reagent . This will use up 3.000 * 4 / 5 = 2.400 (mol) of NH3 and produce 2.400 (mol) of NO.

from A) The molecular mass of NO is 30.01(g/mol), so the mass of NO produced is:
2.400 (mol) x 30.01 (g/mol) = 72.02 (g) of NO.

Answer 2

Atomic weights: N=14 O=16 H=1 NH3=17 O2=32 NO=30 H2O=18

A. 15.00gNO x 1molNO/30gNO x 1molO/1molNO = 0.5 moles O (that's the answer in moles as you specified)

0.5molO x 6.02x10^23atomsO/1molO = 3.01 x 10^23 atoms O

B. 8.517gNH3 x 1molNH3/17gNH3 x 5molO2/4molNH3 x 32gO2/1molO2 = 20.04g O2

C. 45.015gNO x 1molNO/30gNO x 6molH2O/4molNO = 2.2508 moles H2O

D. You work this problem out twice, first how many moles NO from the NH3, then how many moles from the O2. Comparing the answers leads you to know whether it's NH3 or O2 that runs out first (the limiting reagent). The final answer is the moles of NO produced from the limiting reagent.