Balance the following equations. (d) UO 2 + HF → UF 4 + H 2 O

Hello everyone in this question asked to balance the following equations by adding the correct coefficients. We'll see you out A ceo and it's gonna yell, See you Plus c. 0. 2. We have copper, oxygen and carbon on both sides. We have one copper over here To oxygen and one carbon. one copper over here, two oxygen. And while carbon everything is balanced, we can add the coefficients one, one, one in one, And for any three and Plus H two L. And this yields N A O H plus NH three. We have sodium, nitrogen, hydrogen and oxygen on both sides. We have three sodium over here, one nitrogen to hydrogen and one oxygen. And we have one sodium over here, one nitrogen for hydrogen and one oxygen since we have three sodium on the reactive side and only one on the product side, we can multiply by three over here and put a three in front of any O. H. So now we're going to have three plus three for a total of six hydrogen for the most part. Factory over here, we're gonna add a three in front of H 20 And get six hydrogen over here. And now we have three oxygen over here And three over here. So we're going to put a one here, three here, three here and one here and for S. T. O. To let's go to building S. E. Oh three. Love selenium and oxygen on both sides. We have one selenium on the reactive side and for oxygen On the product side we have one selenium and three oxygen Since we have four auction on the reactive side and three on the product side, We need to find the lowest common multiple, which is going to be 12. I hope you must have about four over here, We're Gonna put a four in front of SC. So now we're gonna have four selenium. So if you put a forefront of S C. 02, we're gonna have four selenium over here, We're gonna have eight Plus two for a total of 10 auction over here. And if you put a two in front of 02, we're gonna have 12 oxygen. So we're gonna put a four here. A two here and a four here. Thanks for watching my video. And I hope it was helpful.

Ch.3 - Mass Relationships in Chemical Reactions

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

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McMurry 8th Edition

Ch.3 - Mass Relationships in Chemical Reactions

Problem 38d

Chapter 3, Problem 38d

Balance the following equations. (d) UO₂ + HF → UF₄ + H₂O

Verified step by step guidance

Identify the elements present in the equation: U, O, H, and F.

Write down the unbalanced equation: UO_2 + HF → UF_4 + H_2O.

Count the number of each type of atom on both sides of the equation.

Balance the uranium (U) atoms first, as there is one U atom on each side.

Balance the fluorine (F) atoms by adjusting the coefficient of HF on the reactant side.

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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. Understanding stoichiometry is essential for balancing equations, as it provides the ratios needed to ensure that all atoms are accounted for in the reaction.

Types of Chemical Reactions

Chemical reactions can be classified into several types, including synthesis, decomposition, single replacement, and double replacement reactions. The reaction between UO2 and HF is a type of synthesis reaction, where two or more reactants combine to form a single product. Recognizing the type of reaction helps in predicting the products and understanding the balancing process.