chem help please?

How many moles of O₂ are needed to react completely with 0.470 mole of C₃H₄?

C₃H₄(g) + 4O₂(g) → 3CO₂(g) + 2H₂O(g)
Mole ratio C₃H₄ : O₂ = 1 : 4

Moles of O₂ = (0.470 mol) × 4 = 1.88 mol

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How many grams of CO₂ are produced from the complete reaction of 76.0 g of C₃H₄? (this is c btw)

Molar mass of C₃H₄ = 12×3 + 1×4 = 40 g/mol
Moles of C₃H₄ = (76.0 g) / (40 g/mol) = 1.9 mol

According to the equation, mole ratio C₃H₄ : CO₂ = 1 : 3
Moles of CO₂ = (1.9 mol) × 3 = 5.7 mol

Molar mass of CO₂ = 12 + 16×2 = 44 g/mol
Mass of CO₂ produced = (5.7 mol) × (44 g/mol) = 251 g

OR:
(76.0 g C₃H₄) × (1 mol C₃H₄ / 40 g C₃H₄) × (3 mol CO₂ / 1 mol C₃H₄) × (44 g CO₂ / 1 mol CO₂)
= 251 g

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If the reaction in part c produces 199 g of CO₂, what is the percent yield of CO₂ for the reaction?

Percent yield of CO₂ = (199/251) × 100% = 79.3%

C3H4(g)+4O2(g)⟶Δ3CO2(g)+2H2O(g)

How many moles of O2 are needed to react completely with 0.470 mole of C3H4?
   0.470 X 4  =  .... moles O2

How many grams of CO2 are produced from the complete reaction of 76.0 g of C3H4? (this is c btw)
   76.0/40.0  X  3*44.0  =  .... g CO2

If the reaction in part c produces 199 g of CO2, what is the percent yield of CO2 for the reaction?
    199 / (answer from C)  X 100 =  .... % 
   
How I do this? Do need to balance it? it will ask for grams too and how you do percent yield?  ...  see above --- equation is balanced as written, just use it