determine how many grams of CO2 are produced by burning 2.43g of C4H10?

Balanced equation for burning of
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Mole ratio C₄H₁₀ : CO₂ = 2 : 8

Moles of C₄H₁₀ burned = (2.43 g) / (58.0 g/mol) = 0.0419 mol
Moles of CO₂ produced = (0.0419 mol) × (8/2) = 0.1676 mol
Mass of CO₂ produced = (0.1676 mol) × (44.0 g/mol) = 7.37 g

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OR:

(2.43 g C₄H₁₀) × (1 mol C₄H₁₀ / 58.0 g C₄H₁₀) × (8 mol CO₂ / 4 mol C₄H₁₀) × (44.0 g CO₂ / 1 mol CO₂)
= 7.37 g CO₂

2.43 g C4H10 / 58.12 g/mol X (8 mol CO2 / 2 mol C4H10) X 444.0 g/mol CO2 = 7.36 g CO2

Butane + Oxygen ➜ Carbon Dioxide + water
2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O
molecular weights
C = 12
H = 1
O = 16
2C₄H₁₀ = 2•58 = 116
13O₂ = 13•32 = 416
8CO₂ = 8•44 = 352
10H₂O = 10•18 = 180
check 116+416 = 352+180 = 532

116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O
2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O

ratio of CO₂ to C₄H₁₀ is 352/116

you have 2.43g
2.43g x 352/116 = 7.37 g