Respuesta :
The complete combustion reaction for C16H34 would be as follows:
C16H34 + 17/2O2 ------> 16CO2 + 17H2O
We are given the amount of oxygen to be used for the reaction. This value would be the stating point for the calculations.
0.50 mol O2 ( 1 mol C16H34 / 17/2 mol O2 ) ( 22.4 L / 1 mol ) = 1.317 L = 1317 mL C16H34
C16H34 + 17/2O2 ------> 16CO2 + 17H2O
We are given the amount of oxygen to be used for the reaction. This value would be the stating point for the calculations.
0.50 mol O2 ( 1 mol C16H34 / 17/2 mol O2 ) ( 22.4 L / 1 mol ) = 1.317 L = 1317 mL C16H34
Answer:
0.56 mL
Explanation:
Let's consider the combustion of C₁₆H₃₄.
C₁₆H₃₄ + 24.5 O₂ → 16 CO₂ + 17 H₂O
The molar ratio of C₁₆H₃₄ to O₂ is 1:24.5. The moles of C₁₆H₃₄ that react with 0.050 mol of O₂ are:
0.050 mol O₂ × (1 mol C₁₆H₃₄ / 24.5 mol O₂) = 0.0020 mol C₁₆H₃₄
The molar mass of C₁₆H₃₄ is 226.44 g/mol. The mass of C₁₆H₃₄ is:
0.0020 mol × (226.44 g/mol) = 0.45 g
The density of C₁₆H₃₄ is 0.80 g/mL. The volume of C₁₆H₃₄ is:
0.45 g × (1 mL / 0.80 g) = 0.56 mL