How many mL of hexadecane (C₁₆H₃₄) are needed to react with 0.050 mol O₂, assuming complete combustion occurs? The density of C₁₆H₃₄ is 0.80 g/mL. A)0.99 mL B)0.37 mL C)0.58 mL D)0.46 mL E)0.89 mL

Question

Quizplus · Accepted Answer

The balanced chemical equation for the combustion of hexadecane is:

C16H34 + 25O2 → 16CO2 + 17H2O

From the equation, we can see that the stoichiometric ratio between hexadecane and O2 is 1:25. Therefore, for every 1 mol of hexadecane, 25 mol of O2 are needed for complete combustion.

In this problem, we are given 0.050 mol of O2, so the amount of hexadecane needed can be calculated as follows:

0.050 mol O2 x (1 mol C16H34 / 25 mol O2) = 0.002 mol C16H34

Next, we can use the density of hexadecane to convert from moles to volume:

0.002 mol C16H34 x (1 L / 1000 mL) x (16 g/mol / 0.80 g/mL) = 0.0588 mL

Rounding to the nearest hundredth gives the answer of 0.58 mL, which is Choice C.

How Many ML of Hexadecane (C16H34) Are Needed to React

How Many ML of Hexadecane (C₁₆H₃₄) Are Needed to React