Question

What is the limiting reactant in the following reaction if 1.63g of NH3 is reacted with 3.47g of Cl2 ?

4 NH3 + 3 Cl2 → 3 NH4Cl + NCl3
a. Cl2
b. NCI3
c. NH4Cl
d. NH3​

Answer 1

B

Step-by-step explanation:

you will have to find the number of moles for each 4Nh3 and 3cl2

4NH3

m= 1.63

mr= 14+1x3= 17

n.of moles = 1.63/17=0.09588

3CL2

m=3.47

mr=35.5x2=71

nof moles = 3.47/71= 0.04887

after getting the number of moles you will divide your answer by its ratio

4NH3 3CL2

0.09588/4=0.02397 0.04887/3= 0.0133

the smaller number will be the limiting reagent while the bigger number is the excess reagent