Question

Liquid octane(CH₂(CH₂) CH3) will react with gaseous oxygen (O₂) to produce gaseous carbon dioxide (CO₂) andgaseous water (H₂O). Suppose 5.71 g of octane is mixed with 5.0 g of oxygen. Calculate the maximum mass of water thatcould be produced by the chemical reaction. Round your answer to 2 significant digits.8oloXS ?Ar8

Answer 1

The mass of water produced during the reaction is 2.0 grams

Step-by-step explanation

Given that;

The mass of octane reacted is 5.71 grams

The mass of oxygen reacted is 5.0 grams

Follow the steps below to find the mass of water produced during the reaction

Step 1; Write a balanced equation of the reaction

Recall, that the major products formed when organic compounds undergo combustion reaction are water and carbondioxide

`\text{ 2C}_8H_(18(l))\text{ + 25O}_(2(g))\text{ }\rightarrow\text{ 8CO}_(2(g))\text{ +18H}_2O_(g))`

Step 2; Find the number of moles of octane and oxygen using the below formula

`\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of octane is 114.23 g/mol and the molar mass of oxygen is 32 g/mol

`\begin{gathered} \text{ For Octane \lparen C}_8H_(18)) \\ \text{ mole = }\frac{\text{ 5.71}}{\text{ 114.23}} \\ \text{ mole = 0.049 mol} \\ \\ \text{ For O}_2\text{ \lparen oxygen\rparen} \\ \text{ mole = }\frac{\text{ 5}}{\text{ 32}} \\ \text{ mole = 0.156 mol} \end{gathered}`

In the above calculations, the mole of octane is 0.049 mole and the number of mole of oxygen is 0.156 mol

Step 3; Find the limiting reactant of the reaction

To determine the limiting reactant of the reaction, divide the number of mole of the reactant by its coefficient.

Recall, that the coefficient of octane is 2 and the coefficient of oxygen is 25

`\begin{gathered} \text{ For octane} \\ \text{ }\frac{\text{ 0.049}}{\text{ 2}}\text{ = 0.0245 mol/wt} \\ \\ \text{ for O}_2 \\ \text{ }(0.156)/(25)\text{ = 0.00624 mol/wt} \end{gathered}`

The limiting reactant is oxygen because it has the least mol/wt

Step 4; Find the number of moles of water using a stoichiometry ratio

25 moles O2 react to give 18 moles H2O

Let x represent the number of moles of H2O

`\begin{gathered} \text{ 25 moles O}_2\text{ }\rightarrow\text{ 18 moles H}_2O \\ \text{ 0.156 mol O}_2\text{ }_\rightarrow\text{ x moles H}_2O \\ \text{ Cross multiply} \\ \text{ 25 moles O}_2\text{ }*\text{ x moles H}_2O\text{ = 18 moles H}_2O\text{ }*\text{ 0.156 mol } \\ \text{ Isolate x moles H}_2O \\ \text{ x moles H}_2O\text{ = }\frac{\text{ 18}*\text{ 0.156 }}{25} \\ \text{ x moles H}_2O\text{ = 0.11232 mol} \end{gathered}`

The number of moles of water is 0.11232 mol

Step 5; Find the mass of water using the below formula

`\begin{gathered} \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = mole }*\text{ molar mass} \end{gathered}`

Recall, that the molar mass of water is 18 g/mol

`\begin{gathered} \text{ mass = mole }*\text{ molar mass} \\ \text{ mass = 0.11232}*\text{ 18} \\ \text{ mass = 2.02176 grams} \\ mass\approx\text{ 2.0 to 2 significant figures} \end{gathered}`

Therefore, the mass of water produced during the reaction is 2.0 grams