Calculate the volume of hydrogen needed to saturate the double bonds in 100 g of linoleic acid (rmm = 280).

the answer is 17.28 dm^3

                     15.99 dm³

Explanation:

The molecular formula of Linoleic acid is C₁₈H₃₂O₂. As shown in attached figure, the Linoleic acid contains two double bonds hence we can say that in the Hydrogenation reaction of Linoleic acid 2 mole of H₂ are added across the two double bonds to make it saturated (Stearic acid).

The balance chemical equation for the hydrogenation of Linoleic acid is as follow:

                            C₁₈H₃₂O₂ + 2 H₂  →  C₁₈H₃₆O₂

Calculating moles of Linoleic acid:

                    Moles = Mass / M/Mass

                    Moles = 100 g / 280 g/mol

                    Moles = 0.357 moles Linoleic acid

Calculating moles of Hydrogen:

According to equation:

1 mole C₁₈H₃₂O₂ required  =  2 moles of H₂

So,

0.357 moles Linoleic acid will require  =  X moles of H₂

Solving for X,

                        X  =  2 moles × 0.357 moles / 1 mole

                        X  =  0.714 moles of H₂

Calculating volume of Hydrogen Gas:

Let us assume that the volume is calculated at standart temperature (273.15 K) and pressure (1 atm). We know that STP,

                         1 mole of gas occupies  =  22.4 dm³ of volume

So,

              0.714 moles of gas will occupy  = X dm³ of volume

Solving for X,

                     X =  22.4 dm³ × 0.714 moles / 1 mole

                     X =  15.99 dm³

Note: As the answer is not 17.28 dm³ so it means that the conditions for gas like temperature and pressure are different and are not mentioned in statement. Hence, if the conditions are known then one can use ideal gas equation i.e. PV=nRT to solve for volume.

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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