Balance the following equations: (a) HClO 4 (aq) + P 4 O 10 (s) → HPO 3 (aq) + Cl 2 O 7 (l) (b) Au 2 S 3 (s + H 2 (g) → Au(s) + H 2 S(g)

Hey everyone. Let's go ahead and balance this chemical reaction first. We're going to need to count the number of atoms we have on both sides. Starting off with our react inside, we have two of iron, three of sulfur and two of hydrogen in our product side. We have one of iron, one of sulfur and two of hydrogen 1st. Let's go ahead and balance out our iron. Now, if we add a coefficient of two prior to our iron in our product side, we end up balancing our irons perfectly. Now let's go ahead and look at our sulfur and our hydrogen. Now in order to balance out our cell furs, we're going to have to add a coefficient of three prior to our hydrogen sulfide. And when we do that, we change the number of sulfur and hydrogen we have in our product side. So we have three of sulfur and six of hydrogen since we have six of hydrogen. Now let's go ahead and look at our react inside and we can fix this by adding a three prior to our hydrogen gas, Which gives us six of hydrogen. So our final balanced chemical reaction is going to be like this. So I hope this made sense. And let us know if you have any questions

Ch.3 - Chemical Reactions and Reaction Stoichiometry

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Brown 14th Edition

Ch.3 - Chemical Reactions and Reaction Stoichiometry

Problem 12a,b

Chapter 3, Problem 12a,b

Balance the following equations:
(a) HClO₄(aq) + P₄O₁₀(s) → HPO₃(aq) + Cl₂O₇(l)
(b) Au₂S₃(s + H₂(g) → Au(s) + H₂S(g)

Verified step by step guidance

Step 1: Identify the reactants and the products in the chemical equation. In this case, the reactants are Au2S3 and H2, and the products are Au and H2S.

Step 2: Balance the equation for one type of atom at a time. Start with the atom that appears in the greatest number of compounds on either side of the equation. In this case, start with Au. There are 2 Au atoms on the reactant side and 1 Au atom on the product side. To balance this, place a coefficient of 2 in front of Au on the product side.

Step 3: Next, balance the S atoms. There are 3 S atoms on the reactant side and 1 S atom on the product side in H2S. To balance this, place a coefficient of 3 in front of H2S on the product side.

Step 4: Finally, balance the H atoms. There are 2 H atoms on the reactant side and 6 H atoms on the product side (3 H2S molecules each with 2 H atoms). To balance this, place a coefficient of 3 in front of H2 on the reactant side.

Step 5: Check your work. The final balanced equation should have the same number of each type of atom on both sides of the equation. If it does not, go back and adjust the coefficients until it does.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms for each element is the same on both sides of the equation. This is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Coefficients are used to adjust the number of molecules or formula units to achieve balance.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a reaction. Understanding stoichiometry is essential for balancing equations, as it provides the ratios needed to ensure that all atoms are accounted for.

Types of Chemical Reactions

Chemical reactions can be classified into several types, including synthesis, decomposition, single replacement, and double replacement. Recognizing the type of reaction can help in predicting the products and understanding how to balance the equation. In the given equation, the reaction involves the formation of gold and hydrogen sulfide from gold sulfide and hydrogen, indicating a type of synthesis or decomposition reaction.