6. The heat of vaporization of a liquid is 1,340 Joules per gram. What is the minimum number of Joules needed
to change 40.0 grams of the liquid to vapor at the
boiling point?
A) 33.
B) 1,340
C) 3,280
D) 53,600
7. How many joules of heat are absorbed to raise the
temperature of 435 grams of water at 1 atm from
25°C to its boiling point, 100.°C?

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