Chemistry, 08.04.2020 00:10, maevemboucher78

Liquid octane (CH3(CH2)6CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). What is the theoretical yield of carbon dioxide formed from the reaction of 8.00 g of octane and 38.9 g of oxygen gas?

Be sure your answer has the correct number of significant digits in it.

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answered: dylan2741

24.70g

Explanation:

We'll begin by writing a balanced equation for the reaction between octane and O2 to produce CO2 and H2O. This is illustrated below:

2CH3(CH2)6CH3 + 25O2 —> 16CO2 + 18H2O

Next, let us obtain the masses of octane and O2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of CH3(CH2)6CH3 = 12 + (3x1) + 6[(12 + (2x1)] + 12 + (3x1) = 12 + 3 + 6[14] + 12 + 3 = 12 + 3 + 84 + 12 + 3 = 114g/mol

Mass of CH3(CH2)6CH3 that reacted from the balanced equation = 2 x 114 = 228g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 that reacted from the balanced equation = 25 x 32 = 800g

Next, let us determine which of the reactant is the limiting reactant. This is illustrated below:

From the balanced equation above,

228g of octane required 800g of oxygen for complete Combustion.

Therefore, 8g of octane will require = (8 x 800)/228 = 28.07g of oxygen.

From the calculations made above we can see clearly that oxygen is the excess reactant as the mass of oxygen required (28.07g) is smaller that what was given(i.e 38.9g) from the question and this implies that there are left over mass of oxygen. Therefore, octane is the limiting reactant.

Now we can obtain the theoretical yield of CO2 as illustrated:

The equation for the reaction is given below:

2CH3(CH2)6CH3 + 25O2 —> 16CO2 + 18H2O

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 produced from the balanced equation = 16x44 = 704g

From the balanced equation above,

228g of octane produced 704g of CO2.

Therefore, 8g of octane will produce = (8 x 704)/228 = 24.70g of CO2.

Therefore the theoretical yield of CO2 is 24.70g

answered: Guest

It would be true sodium reacts with water in a violent way

answered: Guest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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Liquid octane (CH3(CH2)6CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (...