73.6k views
3 votes
How many molecules (not moles) of NH3 are produced from 5.01×10−4 g of H2?

User BlueDog
by
8.0k points

2 Answers

2 votes
The chemical formula of ammonia is written as NH3 we can just use the ratio of the atoms since we are not given the reaction. We also make use the avogadro's number to convert from moles to molecules. We calculate as follows:

5.01 x 10^-4 g H2 ( 1 mol / 2.02 g H2 ) ( 2 mol NH3 / 3 mol H2 ) ( 6.022x10
^23 molecules / 1 mol ) = 9.96 x 10^19 molecules NH3
User Nazreen
by
8.1k points
7 votes

Answer : The number of molecules of
NH_3 produced are
10.06* 10^(19)

Explanation :

The given balanced chemical reaction will be,


N_2+3H_2\rightarrow 2NH_3

First we have to calculate the moles of
H_2.


\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=(5.01* 10^(-4)g)/(2g/mole)=2.50* 10^(-4)moles

Now we have to calculate the moles of
NH_3.

From the balanced chemical reaction we conclude that,

As, 3 moles of
H_2 react to produce 2 moles of
NH_3

So,
2.50* 10^(-4)moles of
H_2 react to produce
(2)/(3)* 2.50* 10^(-4)=1.67* 10^(-4)moles moles of
NH_3

Now we have to calculate the number of molecules of
NH_3.

As, 1 mole of
NH_3 produces
6.022* 10^(23) molecule of
NH_3

So,
1.67* 10^(-4)moles of
NH_3 produces
(1.67* 10^(-4))* (6.022* 10^(23))=10.06* 10^(19) molecule of
NH_3

Therefore, the number of molecules of
NH_3 produced are
10.06* 10^(19)

User Jourmand
by
8.2k points