Write equations describing the reactions of Ga with each of the following: F2, O2, S8, and HCl

Name: Problem 25, Chapter 18: Chemical Principle
Uploaded: 2020-08-10T19:29:38-08:00
Duration: 7 min 8 s
Channel: Bin Chen
Description: Write equations describing the reactions of Ga with each of the following: F2, O2, S8, and HCl

Write equations describing the reactions of Ga with each of...

Key Concepts

Chemical Reaction Equations

Chemical reaction equations are symbolic representations of chemical processes where reactants are transformed into products. They involve writing the formulas of the substances involved and balancing the number of atoms of each element on both sides to respect the law of conservation of mass. This concept is fundamental for understanding how substances interact and change during a reaction.

Stoichiometry

Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It involves using balanced equations to determine the proportions of substances involved, ensuring that atoms and charge are conserved. This concept is crucial for accurately predicting the amounts of products formed or reactants needed in a reaction.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between chemical species, resulting in changes in their oxidation states. Understanding these electron transfers is key to explaining why certain reactions occur, especially when metals react with nonmetals or acids, as the metal typically loses electrons (oxidizes) while another species gains electrons (reduces).

Metal Reactivity with Nonmetals and Acids

This concept covers how metals interact with nonmetals such as halogens, oxygen, and sulfur, as well as with acids. Metals often react to form ionic compounds by donating electrons to the nonmetals, leading to products like metal halides, oxides, or sulfides. Reaction with acids typically involves the formation of salts and the release of hydrogen gas, demonstrating the characteristic behavior of metals in different chemical environments.

Transcript

00:01 Okay, so for this question, they want us to write the equation for the reaction of gallium with fluorine, oxygen, sulfur, and hcl.

00:13 So let's start with fluorine.

00:16 So gallium, gaia plus f2 yields gaf3.

00:30 Right so the reason why there's three on fluorine is because the oxidation state of gallium is three and the oxidation state of is plus three and the oxidation state of fluorine is negative one right so this is how we write it so since it's plus three and this is negative one the one goes here and the three goes here and we don't write the one because we know gallium is just writing gallium by itself is one so let's balance it out so we have three fluorines on the right we have two flooring on the left so one side is our number the other side is even number so we have to multiply the right side by two so we can make this three into an even number because no matter one number we put in front of this flooring if we put like a three in front it'll be a six if we put it five in front it'll be a 10 that's always going to be even so we have to make the right even right so two times three is six right we so we need to make the left side six if we put a 3 here, 3 times 2 is 6.

01:36 Now let's get gallium.

01:37 We have 2 on the right, and we only have 1 on the left.

01:40 So we need to balance it out by adding a 2 here, and we will get 2 on both sides.

01:48 Now for the next part, it's gallium with oxygen.

01:58 Use ga203, right? oxidation state of oxygen is negative 2.

02:02 Oxidation state of gallium is plus 3.

02:06 And you get ga2.

02:10 Let's balance it out we have oxygen on the right side so we have three of them and we have two oxygens on the left again we can only get an even number on the left side of oxygen so we need to make the right side even and we can do that just by multiplying it by two because two times any number is even right so we get six and we put a three here we get six on both sides now let's look a gallium so we have four gallium because we have two times three two times two right but we only have one on the lot so if we put a four here we will get a balance gallium now for c they want us to react gallium with sofer and it forms g a 2 s3 okay let's bounce it out so far we have three on the right we have eight on the left oops and we have eight on the left so if we were to multiply the right side by 2 we will get 6 so first but we have 8 on the left so we need more if we multiply it by 3 we get 9 and that is an on number and we can't have that because we can only get an even number on the left side okay so let's change this into a 4 if we get a 4 we have 4 times 12 we get 12 and again we can't get 12 on the left side because if we put a 2 in front of the s 8 we get 16 right so what we can do is is let's increase this to six if we increase this to six we'll have 18 and we can't get 18 on the left side because it's only 8 16 or 24 so let's put an 8 in front of here right so three times 8 is 24 and we can get 24 on the left side if we just add a 3 here right it's because 3 times 8 it goes 24 so now sofer is balanced now we got to balance out the gallium so we have 2 times it by 8, so that's 16.

04:41 We only have one on the left so far, so if we put 8 in front of the gallium, we get 16 on both sides.

04:54 Here for the last part, it's gallium with hcl, and this one is g .a .c .l 3 plus h2 gas.

05:09 This is aqueous.

05:12 Okay, so let's balance out the chlorine for a sec so you have three on the right we have one on the left so we need to bounce it out by adding a three now let's balance out the hydrogens so we have two hydrogens on the right we have three hydrogens on the left all right so in order to balance it out we need to add more onto the right side but if we add any number in front of the h2 we'll always get an even number but on the left side, we can get an or even number, right? so let's just add two here...

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