The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O

The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.

How many moles of CO2 are produced when 35.0 mol C2H2 react completely?
67.2
ANSWER⇒ 70.0 mol CO2
How many moles of CO2 are produced when 84.0 mol O2 completely react?
ANSWER:67.2 mol CO2
Which is the limiting reactant?
ANSWER:oxygen (O2)

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.