Question

balance the following equation: n-octene (C8H16) + potassium permanganate + sulfuric acid yields carbon dioxide+ water+manganese 2 sulfate potassium sulfate. if 100 grams of c8h16 reacts with 150 grams of kmno4 and 400 grams of sulfuric acid, how many grams of carbon dioxide would be formed at 79.5% yield. of the excess reactants, how much of each is left over after the reaction.

Answer 1

The unbalanced chemical equation can be written as:

Now, we can strip the spectator ions such as to write the following ionic equation.

Hence, the two half reactions we can write are

We will balance the half reactions as follows:

We balance the number of C atoms by multiplying 8 on CO₂.

Next, we can add 16 H₂O on left to balance the number of O atoms.

Next we will add 32+16 = 48 H⁺ on right to balance the number of H on both sides.

Next, we will add 48 electrons on right side to balance the number of charges on both side.

Hence, the balanced oxidation half reaction is

Now, we can balance the reduction half as follows:

We balance the number of O atoms by adding 4H₂O on right.

Next, we balance the number of H atoms on both side by adding 8 H⁺ on left.

Now, we balance the number of charges by adding 5 electrons on left.

Hence, the balanced reduction half reaction is

Now, to cancel the electrons involved in both the half reactions while adding, we must cross multiply the number of electrons as follows:

and

Now, we can add both the half reactions as follows:

Now, we can subtract 80 H₂O and 240 H⁺ from both sides of the equation to get the final balanced equation as follows:

Note: Now we can add the spectator ions to complete the balanced equation.

Note that 144 H⁺ on left will be equivalent to 72 H₂SO_4.

Hence, adding K⁺ to MnO₄^- and SO₄^2- to Mn²⁺, we get the following:

Note that we have extra 48 K ⁺ units on the left and 24 SO₄^2- unit on the left. Hence, we have to add 24 K₂SO₄ on the right as required.

Hence, the final balanced reaction will be

Amount of C₈H₁₆ taken = 100 g

Molar mass of C8H16 = 112.24 g/mol

Hence, number of moles of n-octene taken is

Amount of KMnO4 taken = 150 g

Molar mass of KMnO4 =158.034 g/mol

Hence, number of moles of KMnO4 taken is

Amount of sulfuric acid taken = 400 g

Molar mass of sulfuric acid = 98.079 g/mol

Hence, number of moles of sulfuric acid taken is

Now, 5 moles of n-octene reacts completely with 48 moles of KMnO4 and 72 moles of H2SO4.

Hence, number of moles of n-octene that reacts with 4.078 mol of H2SO4 is

Hence, clearly H₂SO₄ is a limiting reagent as we will have some octene left over after reaction.

Now, number of moles of octene that will react with 0.949 mol of KMnO4 is

Hence, clearly KMnO4 is also a limiting reagent.

Hence, Only 0.0988 mol of C8H16 will be able to react because of limiting reagent KMnO4.

5 mol of octene results in 40 moles of CO2.

Hence, number of moles of CO2 that can form from 0.0988 mol of C8H16 is

Molar mass of CO2 = 44.01 g/mol

Given that the percent yield is 79.5 %, the amount of CO2 that can form in the reaction is

Hence, 27.7 g of CO₂ will be formed.

Note that among the reactants, some of octene and H2SO4 will be leftover after the reaction.

Number of moles of octene left over = 0.891 mol - 0.0988 mol = 0.7922 mol

Hence, the mass of octene left over can be calculated as

Now, number of moles of H2SO4 that has reacted with 0.0988 mol of octene is

Hence, number of moles of H₂SO₄ that is left over = 4.078 mol - 1.423 mol = 2.655 mol

Hence, mass of H2SO4 left over after the reaction is