Question

Consider the following reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) If a container were to have 10 molecules of O2 and 52 molecules of NH3 initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Answer 1

Answer:

64 molecules

Explanation:

we have the reaction  

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

That is a balanced equation  

We can see here that we need 5 molecules of O2 for every 4 molecules of NH3  

we have in the beginning 10 molecules of O2 and 52 molecules of NH3

we can say:  

if 4 molecules of NH3 need 5 molecules of O2 how many molecules are needed for 52 molecules of NH3

?

4 molecules of NH3 →5 molecules of O2

52 molecules of NH3 →X molecules of O2

$X=\frac{52 molecules-of- NH_3* 5 molecules-of-O_2}{4 molecules-of-NH_3}=65 molecules-of- O_2$

Since we just have 10 molecules of O2 it is not enough so this is the limiting reagent and all the calculations have to be done base on it.

We use simple rule of three to get all the molecules.

5 molecules of O_2 → 4 molecules of NH_3

10 molecules of O_2 →X molecules of NH_3  

$\frac{4 molecules-of- NH_3* 10 molecules-of-O_2}{5 molecules-of-O_2}=8 molecules-of- NH_3$

Which means that we use all the O2 but just 8 molecules of NH3 and we have 44 less.

NO produced

5 molecules of O_2 → 4 molecules of NO

10 molecules of O_2 → X molecules of NO  

$\frac{4 molecules-of- NO* 10 molecules-of-O_2}{5 molecules-of-O_2}=8 molecules-of- NO$

H2O produced

5 molecules of O_2 → 6 molecules of H_2 O

10 molecules of O_2 → X molecules ofH_2 O  

$\frac{4 molecules-of- H_2 O* 10 molecules-of-O_2}{5 molecules-of-O_2}=12 molecules-of- H_2 O$

Total molecules

44 molecules of NH3 + 8 molecules of NO + 12 molecules of H2O = 64 molecules.