Question

PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will produce ammonia (NH3). Hydrogen and nitrogen gases are reacted to produce the ammonia. For the first batch of ammonia production, 475 g of nitrogen is reacted with excess hydrogen, and 397 g of ammonia are produced.

• Write the balanced equation for the formation of ammonia from hydrogen and nitrogen.

`3Hx_(2) + Nx_(2) --\ \textgreater \ 2NHx_(3)`2NH3

• Calculate the theoretical yield of ammonia. Work must be shown to earn credit.



• Calculate the percent yield for the ammonia production. Work must be shown to earn credit.

Answer 1

Part 1 : Balanced reaction,
`3H_2(g)+N_2(g)\rightarrow 2NH_3(g)`

Part 2 : The theoretical yield of
`NH_3` gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of
`N_2` = 475 g

Molar mass of
`N_2` = 28 g/mole

Molar mass of
`NH_3` = 17 g/mole

Experimental yield of
`NH_3` = 397 g

Answer for Part (1) :

The balanced chemical reaction is,

`3H_2(g)+N_2(g)\rightarrow 2NH_3(g)`

Answer for Part (2) :

First we have to calculate the moles of
`N_2`.

`\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=(475g)/(28g/mole)=16.96moles`

From the given reaction, we conclude that

1 moles of
`N_2` gas react to give 2 moles of
`NH_3` gas

16.96 moles of
`N_2` gas react to give
`(2)/(1)* 16.96=33.92` moles of
`NH_3` gas

Now we have to calculate the mass of
`NH_3` gas.

`\text{ Mass of }NH_3=\text{ Moles of }NH_3* \text{ Molar mass of }NH_3`

`\text{ Mass of }NH_3=(33.92moles)* (17g/mole)=440.96g`

Therefore, the theoretical yield of
`NH_3` gas = 440.96 g

Answer for Part (3) :

Formula used for percent yield :

`\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}* 100`

`\% \text{ yield of }NH_3=(397g)/(440.96g)* 100=90.03\%`

Therefore, the % yield of ammonia is 90.03 %