How many liters of ch4(g), measured at 22.6 ∘c and 768 mmhg , must be burned to provide the heat needed to vaporize 3.82 l of water at 100 ∘c? δcombh=−8.90×102kjmol−1 ch4. for h2o(l) at 100 ∘c, d=0.958gcm−3 and δvaph=40.7kjmol−1.
express your answer in liters to three significant figures.

i have a feeling this is a gay person trolling to make straights look bad.

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