A3.35cm air bubble forms in a deep lake at a depth where the temperature is 19.2 degrees celcuis and the pressure is 1564mmhg. the bubble rises to a depth where the temperatuer is now 26.5 degrees celsius and the pressure is 825.5 mmhg. assuming the amount of air in the buble has not changed, calculate its new volume

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

c. is the answer

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