Balance each chemical equation. a. Na 2 S(aq) + Cu(NO 3 ) 2 (aq) → NaNO 3 (aq) + CuS(s) b. N 2 H 4 (l) → NH 3 (g) + N 2 (g) c. HCl(aq) + O 2 (g) → H 2 O(l) + Cl 2 (g) d. FeS(s) + HCl(aq) → FeCl 2 (aq) + H 2 S(g)

Hi everyone here, we have a question asking us to balance a fallen chemical equation cobalt nitrate plus ammonium carbonate forms, cobalt carbonate plus ammonium nitrate. So let's rewrite this. Now let's count what we have on both sides. So on our reactant side we have one cobalt we have three plus two nitrogen, so five We have three times three, so nine oxygen plus three more, so 12 oxygen. And we have four times two hydrogen, so eight hydrogen. And lastly we have one carbon. And on our product side we have to cobalt We have two nitrogen. We have three times three oxygen, so nine oxygen plus three more. So 12 oxygen. We have four hydrogen And we have three carbon. So first we're going to balance our cobalt and to do that, we're going to put a two here. So now we have two cobalt and we have two times three. So six nitrogen plus two, so eight nitrogen. And we have six Times three oxygen. So 18 oxygen. Now we're going to balance our carbon. So we're going to put a three here. So we have three times two nitrogen. So that's six plus two times three Nitrogen. So that's six. So we have 12 nitrogen And we have three Times four times 2 hydrogen. So we have 24 hydrogen And we have three carbon. We have three times three, so nine plus two times three times three. So We have 27 oxygen. And now we need to balance our nitrogen. So we're going to put a six in front of our ammonium nitrate. So we have six Plus six nitrogen, so we have 12 nitrogen. We have six times 4 hydrogen, So 24 hydrogen. We have six times three oxygen, plus three times three oxygen, So 27 oxygen. And now we are all balanced. So that is our final balanced chemical equation. Thank you for watching. Bye.

Ch.3 - Molecules, Compounds & Chemical Equations

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Tro 4th Edition

Ch.3 - Molecules, Compounds & Chemical Equations

Problem 112

Chapter 3, Problem 112

Balance each chemical equation. a. Na₂S(aq) + Cu(NO₃)₂(aq) → NaNO₃(aq) + CuS(s) b. N₂H₄(l) → NH₃(g) + N₂(g) c. HCl(aq) + O₂(g) → H₂O(l) + Cl₂(g) d. FeS(s) + HCl(aq) → FeCl₂(aq) + H₂S(g)

Verified step by step guidance

Step 1: For equation (a), start by balancing the number of sodium (Na) atoms. There are 2 Na atoms in Na2S, so ensure there are 2 Na atoms in NaNO3 by placing a coefficient of 2 in front of NaNO3.

Step 2: Next, balance the sulfur (S) atoms in equation (a). There is 1 S atom in both Na2S and CuS, so they are already balanced.

Step 3: Balance the copper (Cu) atoms in equation (a). There is 1 Cu atom in both Cu(NO3)2 and CuS, so they are balanced.

Step 4: Balance the nitrate (NO3) groups in equation (a). There are 2 NO3 groups in Cu(NO3)2, so ensure there are 2 NO3 groups in NaNO3 by placing a coefficient of 2 in front of NaNO3.

Step 5: For equation (b), balance the nitrogen (N) atoms. There are 2 N atoms in N2H4, so ensure there are 2 N atoms in the products by having 1 N2 molecule and 2 NH3 molecules.

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Key Concepts

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

Law of Conservation of Mass

The Law of Conservation of Mass states that matter cannot be created or destroyed in a chemical reaction. This principle is fundamental when balancing chemical equations, as it requires that the number of atoms of each element must be the same on both sides of the equation. Understanding this law helps ensure that all reactants and products are accounted for in the balanced equation.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. It involves using mole ratios derived from the coefficients of the balanced equation to determine the amounts of substances involved. Mastery of stoichiometry is essential for accurately balancing equations and predicting the outcomes of chemical reactions.

Types of Chemical Reactions

Recognizing the types of chemical reactions—such as synthesis, decomposition, single replacement, and double replacement—can aid in balancing equations. Each type has specific patterns and rules that can simplify the balancing process. Understanding these categories allows for a more strategic approach to identifying how reactants transform into products.