Ammonia gas can be prepared by the reaction CaO ( s ) + 2 NH 4 Cl ( s ) ⟶ 2 NH 3 ( g ) + H 2 O ( g ) + CaCl 2 ( s ) In an experiment, 27.6 g of ammonia gas, NH 3 , is produced when it was predicted that 50.2 g NH 3 would form.

The given question is incomplete. The complete question is ;

Ammonia gas can be prepared by the following reaction:

In an experiment, 27.6 g of ammonia gas, , is produced when it was predicted that 50.2 g of would form.

What is the theoretical yield of ? What is the actual yeild of ? What is the percent yeild of ?

a) 50.2 g

b) 27.6 g

c) 5.0 %

Explanation:

a) Theoretical yield is defined as the amount of the product that is possible in the reaction.

Theoretical yield of = 50.2 g

b) Experimental or actual yield is defined as the amount of the product that is actually formed in the reaction.

Experimental yield of = 27.6 g

c) Percentage yield is defined as the ratio of experimental yileld to the theoretical yield in terms of percentage.

Hence percentage yield for the reaction is 55.0 %.

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.