Question

2 C4H10 + 13 O2--> 8 CO2 + 10 H2O

How many molecules of CO2 are produced from the combustion of 100 grams of butane (C4H10)?

Answer 1

4.14 x 10²⁴ molecules CO₂

Step-by-step explanation:

2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O

To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.

4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀

100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀

6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂