In the presence of H2 and Pt, ethylene is converted to ethane. H2C=CH2 + H2 (ethylene) >> CH3CH3 (ethane)

a. What is the theoretical yield of ethane (in moles) if 4.00 mol of ethylene are reacted with 2.00 mol of H2?
Answer: 2 mol
b. What is the theoretical yield if 50.0 g of ethylene are reacted with 25.0 g of H2?

H2C=CH2 + H2 = > CH3CH3

since the equation is balanced when 4 moles of ethene react with 2 moles of H2 the ethene is the limiting reagent and 2 moles of ethane are formed

moles of ethene in 50g = 50 / 28 = 1.786 moles . these will react with 1.786 moles of H2
but we have 25 / 2 moles of H2 so the ethene is the limiting reagent

and the theoretical yield of ethane is 1.786 moles or 1.786 x 30 g = 53.58g

I hope you understand the reasoning ...