Estimate the maximum volume of alkene that should be in the distillate when thereaction is completed. Use 0.80 g/mL as the density of the alkene mixture.
we have 18.42ml or 14.43 g of methylcyclohexanol.



Answer :

The maximum volume of alkane present in the distillate when the reaction is completed is 18.04 mL of alkane.

When the distillation of methylcyclohexanol is done, the product formed after the distillation is

C₇H₁₄O→C₇H₁₂+H₂O

To find out the volume of alkene mixture, we need to use the density formula which is given as

density=mass/volume

Rearrange the formula for volume

volume=mass/density

We have mass of methylcyclohexanol, which is 14.43g and density is 0.80 g/mL. Plug values in the formula

volume=(14.43g×mL/0.80g)

volume=18.0375 mL

volume=18.04 mL

Therefore, the volume of alkane present in the distillate when the reaction of methylcyclohexanol is completed is 18.04 mL.

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