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Model Write the mole ratios for the reaction of hydrogen gas and oxygen gas, \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} 0 .\) Make a sketch of six hydrogen molecules reacting with the correct number of oxygen molecules. Show the water molecules produced.

Short Answer

Expert verified
The mole ratios for the reaction are 2:1:2 (H\(_2\):O\(_2\):H\(_2\)O). With 6 H\(_2\) molecules, use 3 O\(_2\) molecules to produce 6 H\(_2\)O molecules.

Step by step solution

01

Understanding the Chemical Equation

The chemical equation given is \(2\mathrm{H}_{2} (g) + \mathrm{O}_2 (g) \rightarrow 2\mathrm{H}_{2}O\). It shows that two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water.
02

Identifying Mole Ratios

Based on the chemical equation, the mole ratio of \(\mathrm{H}_2\) to \(\mathrm{O}_2\) to \(\mathrm{H}_2O\) is 2:1:2. This means for every 2 moles (or molecules) of hydrogen, 1 mole (or molecule) of oxygen is required to form 2 moles (or molecules) of water.
03

Using Mole Ratios in Sketch

To sketch six hydrogen molecules reacting with oxygen, use the mole ratio. Six hydrogen molecules mean using three times the basic ratio of 2:1:2. Therefore, you'll need 3 molecules of \(\mathrm{O}_2\) (since 6 hydrogen molecules equate to \(2 \times 3\) moles of H2), and this will produce six molecules of water.
04

Creating the Sketch

Draw six \(\mathrm{H}_2\) molecules. According to the ratio, add three \(\mathrm{O}_2\) molecules. After the reaction, draw six \(\mathrm{H}_2O\) molecules as products. Remember, each water molecule consists of two hydrogen atoms and one oxygen atom.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equation Interpretation
Interpreting a chemical equation involves understanding the symbolic representation of a chemical reaction. It shows the reactants and products involved, along with their quantities. For the reaction of hydrogen and oxygen, the equation is:\[ 2 \mathrm{H}_{2} (g) + \mathrm{O}_2 (g) \rightarrow 2 \mathrm{H}_{2}O \]This equation communicates several important details:
  • The reactants are hydrogen gas \(\mathrm{H}_2\) and oxygen gas \(\mathrm{O}_2\).
  • The product is water \(\mathrm{H}_2O\).
  • Coefficients (numbers before formulas) indicate the number of molecules or moles involved in the reaction: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to form 2 molecules of water.
Understanding these basics helps us predict how substances interact and the quantities needed for reactions. Break the equation into components to deeply grasp what it's communicating.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships of the components in a chemical reaction. It helps us understand how much of each substance we need and what will be produced in a reaction. In the given equation:\[ 2 \mathrm{H}_{2} (g) + \mathrm{O}_2 (g) \rightarrow 2 \mathrm{H}_{2}O \]We can identify a mole ratio of 2:1:2.

Practical Use of Mole Ratios

In practice, this means:
  • For every 2 moles of hydrogen \(\mathrm{H}_2\), 1 mole of oxygen \(\mathrm{O}_2\) is required.
  • These will react to produce 2 moles of water \(\mathrm{H}_2O\).
If you start with 6 moles of hydrogen, for example, you'll need 3 moles of oxygen, producing 6 moles of water. Grasping stoichiometry is crucial as it lets us scale reactions up or down based on our needs.
Always check if your quantities align with the mole ratios to ensure the accuracy of your calculations.
Hydrogen and Oxygen Reaction
The reaction between hydrogen and oxygen to form water is a fundamental process in chemistry. Not only is it important for understanding stoichiometry and chemical equations, but it also has practical real-world applications. When hydrogen \(\mathrm{H}_2\) and oxygen \(\mathrm{O}_2\) gases combine:\[ 2 \mathrm{H}_{2} (g) + \mathrm{O}_2 (g) \rightarrow 2 \mathrm{H}_{2}O \]They form water through a highly exothermic reaction.

Visualizing the Reaction

Suppose we start with 6 hydrogen molecules:
  • This requires 3 oxygen molecules to adhere to the 2:1:2 ratio.
  • After the reaction, we obtain 6 water molecules.
Visualizing this:- Start with groups of hydrogen and oxygen molecules.- After combining, each pair of \(\mathrm{H}_2\) joins with a single \(\mathrm{O}_2\) to form water.This basic reaction underpins the combustion engines and many energy-generating processes, making it essential knowledge for everyday chemical applications.

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Most popular questions from this chapter

Gasohol is a mixture of ethanol and gasoline. Balance the equation, and determine the mass of \(\mathrm{CO}_{2}\) produced from the combustion of 100.0 g of ethanol. \begin{equation} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \end{equation}

Explain why a balanced chemical equation is needed to solve a stoichiometric problem.

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right),\) also known as grain alcohol, can be made from the fermentation of sugar \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right) .\)The unbalanced chemical equation for the reaction is shown below. \begin{equation} -\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightarrow_{-} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+_{-} \mathrm{CO}_{2} \end{equation} \begin{equation} \begin{array}{l}{\text { Balance the chemical equation and determine the mass }} \\ {\text { of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \text { produced from } 750 \mathrm{g} \text { of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \text { . }}\end{array} \end{equation}

Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosilicic acid \(\left(\mathrm{H}_{2} \mathrm{SiF}_{6}\right)\) \begin{equation} \mathrm{SiO}_{2}(\mathrm{s})+6 \mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \end{equation} 40.0 \(\mathrm{g} \mathrm{SiO}_{2}\) and 40.0 \(\mathrm{g}\) HF react to yield 45.8 \(\mathrm{g} \mathrm{H}_{2} \mathrm{SiF}_{6}\) \begin{equation} \begin{array}{l}{\text { a. What is the limiting reactant? }} \\ {\text { b. What is the mass of the excess reactant? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. What is the theoretical yield of } \mathrm{H}_{2} \mathrm{SiF}_{6} \text { ? }} \\ {\text { d. What is the percent yield? }}\end{array} \end{equation}

When 9.59 g of a certain vanadium oxide is heated in the presence of hydrogen, water and a new oxide of vanadium are formed. This new vanadium oxide has a mass of 8.76 g. When the second vanadium oxide undergoes additional heating in the presence of hydrogen, 5.38 g of vanadium metal forms. \begin{equation} \begin{array}{l}{\text { a. Determine the empirical formulas for the two }} \\\ {\text { vanadium oxides. }} \\ {\text { b. Write balanced equations for the steps of the reaction. }} \\ {\text { c. Determine the mass of hydrogen needed to complete }} \\ {\text { the steps of this reaction. }}\end{array} \end{equation}

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