Potassium superoxide (KO2, 71.10 g/mol) can be used to generate oxygen gas. What is the theoretical yield (accepted) of O2 (32.00 g/mol) if 500.0 g KO2 reacts with excess H2O (18.02 g/mol)?

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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

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