When 9.59 g of a certain vanadium oxide is heated in the presence of hydrogen, water and a new o

When 9.59 g of a certain vanadium oxide is heated in the...

Key Concepts

Role of Hydrogen as a Reducing Agent

Hydrogen acts as a reducing agent by donating electrons during a chemical reaction, which results in the reduction of another species, such as a metal oxide. In such reactions, hydrogen itself is oxidized, typically forming water. Recognizing the role of hydrogen is important for understanding both the mechanism of the reduction process and the stoichiometric relationships in reactions where it is involved.

Balanced Chemical Equations

Writing balanced chemical equations requires ensuring that the number of atoms for each element and the overall charge are the same on the reactant and product sides, in accordance with the law of conservation of mass. This process involves correctly assigning coefficients to substances in the equation, which is particularly important in reactions where multiple steps or redox processes occur.

Empirical Formula Determination

This concept involves finding the simplest whole-number ratio of atoms in a compound based on mass data. When given masses of different elements in a compound or reaction, the objective is to convert these masses to moles using their atomic masses, then determine the simplest ratio that defines the compound's empirical formula. It is fundamental to understanding the composition of a substance in chemical analysis.

Stoichiometry

Stoichiometry is the quantitative aspect of chemistry that deals with the relationships between reactants and products in a chemical reaction. By converting given masses to moles, applying mole ratios from balanced equations, and converting back to mass if necessary, one ensures that the law of conservation of mass is adhered to. It is crucial for determining how much product is produced or how much reactant is needed in a reaction.

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between species, resulting in changes in the oxidation states of elements. In this context, hydrogen serves as a reducing agent, meaning it donates electrons to another species (such as a metal oxide), reducing it while being oxidized itself. Understanding redox processes is essential for analyzing reactions where substances undergo a change in oxidation state.

Transcript

00:01 So for part a of this problem, we need to find the empirical formula or lowest whole number ratio for each of these different compounds of vanadium oxide.

00:13 Now we know that there's 5 .38 grams of vanadium if the oxygen is gone.

00:20 With that information, we can determine that the mass of oxygen in this first compound, if we take 9 .59 to subtract 5 .38, grams vanadium from it were left with the 4 .21 grams of oxygen and then here we just do 8 .76 grams minus 5 .38 to get 3 .38 grams of oxygen.

00:51 Now we can't find a ratio here until we get everything converted into moles.

00:58 So for every mole of vanadium if you look on the periodic table it's 50 .9 grams.

01:09 I round it to the nearest tenth.

01:10 It's always good to do that.

01:14 That'll give us what we need for moles.

01:16 You just divide it.

01:23 0 .106 moles.

01:27 So we can use that for both compounds.

01:30 It should be the same.

01:31 Now oxygen here is 4 .21 grams, divided by 16 grams of oxygen for every one mole.

01:43 And that will give you 0 .263.

01:50 Do it with this one, 3 .38 grams divided by 16.

01:59 And you'll get 0 .211.

02:05 Now you have everything in moles.

02:07 All they have to do is compare the mole numbers to each other.

02:12 For the first compound, i'm comparing 0 .263 moles of oxygen to 0 .106 moles of vanadium.

02:32 If i divide those, i'll find out how many moles of oxygen there are for every one mole of vanadium.

02:38 And if you do that, you get a number very close to 2 .5 for the oxygen and forever.

02:51 One vanadium.

02:52 Now that's not a whole number ratio.

02:54 So it's the lowest ratio, but it's not whole number.

02:58 So you got to multiply the top and bottom by two and you get five to two, which means it's v205.

03:12 So your answer here is v205.

03:18 Now what about the other compound? now i've got the zero 0 .211 over again 0 .106.

03:31 And that's, remember, that's oxygen, and that's for every vanadium.

03:36 Do that.

03:37 You should get it over very close to two, which means it's two oxygens for every one vanadium.

03:43 You always got to write the metal first.

03:45 V -o -2.

03:47 So that's your second one.

03:50 So those are the two empirical formulas or lowest whole -number ratios for part a.

03:54 Now for part b, i can use these empirical formulas to complete the balanced equation.

04:03 So we know from the question that we combined each one with hydrogen gas.

04:08 Remember, hydrogen gas is diatomic.

04:15 And we know water and was one of the products.

04:18 But i know the second compound of vanadium oxide was part of the product.

04:26 And then i'll put water here.

04:30 Down here, it was just vanadium, but pure vanadium metal and h2o.

04:40 Now, i'm going to put, you want to have space in front of each one, because you might have to put coefficient numbers.

04:52 So look at the first reaction.

04:57 Just start with v.

04:59 I got v2 here.

05:01 So there's two vs, and on the right there's only one v.

05:04 So i'll start by putting a two in front of here...

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