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


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

Sagot :

CrystalQueen

Answer:

4.14 x 10²⁴ molecules CO₂

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

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