How many moles of pbr3 contain 3.68 ã 1025 bromine atoms?

Given data:

# atoms of Br in PBr3 = 3.68 * 10^25

Explanation:

In PBr3,

1 mole of PBr3 contains 3 moles of Br

Now,

1 mole of Br contains Avogadro number i.e. 6.023 * 10^23 atoms of Br

Therefore, the # moles of Br corresponding to 3.68*10^25 atoms would be:

= 3.68 * 10^25 atoms * 1 mole Br/6.023*10^23 atoms = 61.099 moles of Br

Since,

3 moles of Br is present in 1 mole of PBr3

then, 61.099 moles of Br would correspond to:

61.099 moles Br * 1 mole PBr3/3 moles Br = 20.366 moles of PBr3

Ans: 20.4 moles of PBr3

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