Calculate the mass of butane needed to produce 88.2 g of carbon dioxide.?

Molar mass of C₄H₁₀ (butane) = (12.0×4 + 1.0×10) = 58.0 g/mol
Molar mass of CO₂ (carbon dioxide) = (12.0 + 16.0×2) = 44.0 g/mol

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

No. of moles of CO₂ formed = (88.2 g) / (44.0 g/mol) = 2.005 mol
No. of moles of C₄H₁₀ needed = (2.005 mol) × (2/8) = 0.5013 mol
Mass of C₄H₁₀ needed = (0.5013 mol) × (58.0 g/mol) = 29.1 g (to 3 sig. fig.)

Mass of butane....

2C4H10(g) + 13O2(g) --> 8CO2(g) + 10H2O(g)
....?g ................ .................88.2g

Simply use the unit-factor method. It will give a one-line setup, with one chain calculation and practically zero chance of error.

88.2g CO2 x (1 mole CO2 / 44.0g CO2) x (1 mol C4H10 / 4 mol CO2) x ( 58.0g C4H10 / 1 mol C4H10) = 29.1g C4H10

...
since the amount of carbon in both will be the same we can use this equation
mass of carbon in butane sample = mass of carbon in CO2 sample
Xg butane * 48 g C / 58 g butane = 88.2 g CO2 * 12 g C / 44 g CO2
X = 29.1 g of butane

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