For each molecules give the following information: a) lewis structure b) electron domain geometry c) molecular geometry (shape) d) bond angles associated with the single bonds (< , =, > the bond angles for the edg) e) polar or nonpolar bonds in the molecule f) polarity of the molecule g) type of hybridization h) number of sigma (σ) and pi (π) bonds in the molecule a. cbr2h2 b. pcl3 c. so4 2− d. bh3 e. co

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

16

explanation:

the marks go by 4s