For the following reaction, 4.30 grams of benzene (CH) are allowed to react with 17.6 grams of oxygen gas. benzene (CH) (1) + oxygen (g) —>carbon dioxide (g) + water (g)
What is the maximum amount of carbon dioxide that can be formed?
grams
What is the FORMULA for the limiting reagent?
grams
What amount of the excess reagent remains after the reaction is complete?

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

i think is conclusions.i think so