The following are artificial nuclear reactions that occur by bombardment. Balance each equation, identifying the missing particles' mass number, atomic number, and element symbol:

238/92 U + 14/7 N ---> _ + 5 1/0 n

252/98 Cf + _ ---> 257/103 Lr + 5 1/0 n

57/24 Cr + 4/2 He ---> _ + 56/26 Fe

40/18 Ar + 4/2 He ---> _ + 1/1 H

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .

it is the most commonly accepted one

it is stable as it is in group 0 which have full outer shells of electrons