1). For the following reaction if we react .1037 g NH 3 with .2297 g Cl 2 , what

Question

1).

For the following reaction if we react .1037 g NH3 with .2297 g Cl2, what is the limiting reactant?

4  NH3  +  3 Cl2  à3  NH4Cl  +    NCl3

2).

For the following reaction if we react .1037 g NH3 with .2297 g Cl2, what is the theoretical yield of  NH4Cl?

4  NH3  +  3 Cl2  à3  NH4Cl  +    NCl3

3).

For the following reaction if we react .1037 g NH3 with .2297 g Cl2,

and only .1473 g of NH4Cl is produced from this reaction, what is the

percent yield of NH4Cl?

4  NH3  +  3 Cl2  à 3  NH4Cl  +    NCl3

Explanation / Answer

4  NH3  +  3 Cl2 -----> 3NH4Cl  +    NCl3

1 Mol 0.75mol .75mol 0.25mol

0.1037/17 0.2297/71

=0.0061   = 0.003235

0.0061 mol of NH3 reacts with 075*0.0061=0.004575 mol of Cl2. Thus Cl2 is limiting agent as 0.004575>0.003235.

2) The theoretical yield of NH4Cl = number of mol of Cl2 consumed= 0.003235mol = 0.003235*53.5 =0.1731 gram.

3)percentage yield = observed wieght of NH4Cl / Calculated NH4Cl Wieght *100 = 0.1473/0.1731 *100 = 85.1 %.

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