Balance the following equations. (b) Ca 3 (PO 4 ) 2 (s) + SiO 2 (s) + C(s) --> P 4 (g) + CaSiO 3 (l) + CO(g)

Hey everyone, we're asked to write a balanced equation by providing the correct coefficients. So let's go ahead and rewrite this equation. We have potassium hydroxide plus iron, Three oxide plus oxygen, we then produce potassium fairy plus our water 1st. Let's go ahead and calculate the number of each element on each side. Starting with our react inside. Looking at our potassium, we have one a potassium. Next looking at our iron, we have two of iron. Looking at our oxygen, we have six of oxygen and looking at our hydrogen, we have one of hydrogen. Next looking at our product side, we have two of potassium, we have one of iron. We have five of oxygen and two of hydrogen 1st. Let's go ahead and balance out our iron on our product side and we can do so by adding a coefficient of two prior to our potassium ferry. So now we have two of iron, We have four a potassium and we have nine of oxygen. Now let's go ahead and balance out our potassium in our react inside. We can go ahead and add a coefficient of four prior to our potassium hydroxide. Once we do that, we get four potassium, we got nine of oxygen and we get four of hydrogen. Next let's go ahead and balance out our hydrogen on our product side and we can do so by adding a two prior to our water. Once we do that, we end up with four of hydrogen, but now our oxygen's will also change. Now we have 10 of oxygen. Looking at the values on our product and our react inside. It looks like our oxygen's are the only difference. Now if we add a coefficient of 3/2 prior to our oxygen, we're able to get those 10 oxygen's and everything seems to be balanced out, but we can't have a fraction as a coefficient. So we will have to multiply our whole equation by two. So once we do that, our balanced chemical equation is going to change. So we're going to end up With eight a potassium hydroxide Plus two of our iron, three oxide Plus three of our oxygen. And our products are going to be four of our potassium ferry, plus four of our water. And this is going to be our balanced chemical equation. Now, I hope that made sense. And let us know if you have any questions.

Ch.3 - Mass Relationships in Chemical Reactions

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McMurry 8th Edition

Ch.3 - Mass Relationships in Chemical Reactions

Problem 43b

Chapter 3, Problem 43b

Balance the following equations. (b) Ca₃(PO₄)₂(s) + SiO₂(s) + C(s) --> P₄(g) + CaSiO₃(l) + CO(g)

Verified step by step guidance

Step 1: Identify the different types of atoms in the equation. In this case, we have Calcium (Ca), Phosphorus (P), Oxygen (O), Silicon (Si), and Carbon (C).

Step 2: Start by balancing the atoms that appear in the most complex molecules first. In this case, we can start with Phosphorus (P). There are 2 P atoms on the left side and 4 on the right side. To balance this, we can put a coefficient of 2 in front of Ca3(PO4)2.

Step 3: Next, balance the Calcium (Ca) atoms. There are 6 Ca atoms on the left side and 1 on the right side. To balance this, we can put a coefficient of 6 in front of CaSiO3.

Step 4: Now, balance the Silicon (Si) atoms. There is 1 Si atom on the left side and 6 on the right side. To balance this, we can put a coefficient of 6 in front of SiO2.

Step 5: Finally, balance the Carbon (C) and Oxygen (O) atoms. There are 8 O atoms on the left side and 18 on the right side, and 1 C atom on the left side and 1 on the right side. To balance this, we can put a coefficient of 10 in front of CO.

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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. To balance an equation, coefficients are adjusted in front of the chemical formulas to achieve equal atom counts.

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 based on balanced equations. Understanding stoichiometry is essential for calculating yields, determining limiting reactants, and performing conversions between moles and grams.

Phases of Matter in Reactions

In chemical equations, the phases of matter (solid, liquid, gas) are indicated to provide context about the state of each substance involved in the reaction. This information is crucial for understanding reaction conditions and mechanisms. For example, solids (s) may require different handling compared to gases (g) or liquids (l), influencing the reaction's feasibility and rate.