Question

Complete the following fission reaction. 239 94 Puton → + 91 38 Sr+35n 146 Ba 56 143 Xe 54 91 Sr 38 148 Ba 56 91 38 Mn

Answer 1

this question can be easily solved by balancing the mass number and atomic number of reactants and products of this reaction,

here reactants are Pu and a neutron and products: unknown element,Sr, 3 neutrons

total mass number of reactants= 239+1 =240 (a nuetron has mass number of 1)

let mass number of unknown element be x

total mass number of products=x+91+(3*1)=x+94

we know that ;total mass number of reactants=total mass number of products

so x+94=240; implies x=240-94=146 ;so mass number of unknown element =146

let atomic number of unknown element be y

total proton number of products=y+38+(3*0)=y+38

total proton number of reactants=94+0=94

total proton number of products=total proton number of reactants

so y+38=94 ; y=94-38=56 atomic number of unknown element is 56

now looking at options

¹⁴⁶Ba₅₆ matches our findings

so ¹⁴⁶Ba₅₆ is the answer

this problem is done based on conservation of mass.

thank you