6

The number of "equivalents" of a reagent refers to how much of the reagent there is relative to another reactant of interest. To use a concrete example, if we are interested in the $\mathrm{S_N2}$ transformation $$\ce{CH3Cl -> CH3Br},$$ a possible way of bringing this about would be to treat the starting material (methyl chloride) with a certain amount of ...


5

First, provided the ball is an atom, you are not writing a chemical equation for a dozen of balls, rather a dozen times at least $10^{16}$ (to speak of equilibrium), usually a dozen times $10^{23}$ (macroscale that normal people operate on). That's not the level you want to use addition for; smaller numbers are easier to operate with. Second, it's not ...


5

In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.


4

Assuming you have an excess of $\ce{KNO3}$, the ratio of $\ce{NaN3}$ consumed to $\ce{N2}$ produced would be 5:8. Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10. By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $\ce{N_2}$ are produced for every ...


4

Before we come to your actual question, we should clarify a few other problems related to quantities and units in your expression: moles = mass / RFM It is essential to distinguish between quantities and units. For example, density is defined as “mass per volume” and not “mass per litre”. In your expression “moles” is a unit name; but it should actually ...


4

The proper stoichiometry may not be the only reason for the friend's success, but also the overall preciseness of their preparation procedure. Chemical reactions are not generally that much critical, but there are lot of cases, that are. There are few aspects to consider: Excessive reagents may affect the effectiveness of the product isolation. ...


4

The question in the book is about elemental composition of the given substance, in terms of mass. You are given the empirical formula, i.e. what could be obtained by elemental analysis, namely what elements composed the substance and in which relative ratio. This does not tell you about the spatial arrangement of the various elements and the way they bound ...


3

The purely combinatorial methods are a first step in modeling chemical reactions. They alone however cannot consider the possibility of the chemical reactions and the chemical stability of the reaction products. $\ $ 1.) You can build all combinations of at least two formulas, all from the original set of given chemical formulas, and treat this as your set ...


3

How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass? You find unbalanced equations as exercises and as a step in writing a balanced equation when there is insufficient information that still needs to be discovered. Balancing equations as an exercise The observable ...


3

You can mention experimental details and how you achieved 1:4 ratio, which may explained if we know the maximum temperature of heating. CuSO4.5H2O loses water of hydration in steps as a function of temperature showing that those five water molecules are not equivalent. In this figure from German Wiki, follow the green line (ignore blue). The y-axis is % ...


3

In combustion the "energy" is really the heat that is released. This heat causes the expansion that moves the piston upwards. However, the fuel and the oxygen did not enter the engine with energy, and then leave without it. The fuel and the oxygen did not leave at all, different molecules did. The heat that was released was due to the reaction that changed ...


3

Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me): $$\ce{$x$ \overset{0}{Zn} + y H\overset{+5}{N}O3 -> $a$ \overset{+2}{Zn}(NO3)2 + b H2O + $c$ \overset{-3}{N}H4NO3}$$ $$ \begin{align} \ce{\overset{+5}{N} + 8 e- &→ \overset{-3}{N}} \tag{red}\\ \ce{\overset{0}{...


3

What is the sum of the coefficients $(a + b + c)$ in the following reaction? $$\ce{x Zn + y HNO3 -> a Zn(NO3)2 + b H2O + c NH4NO3}$$ This is a redox reaction. During OP's comments, it was clear that he/her got the correct answer, but it wasn't clear that OP understand this is a redox reaction. To balance the equation, it may be better work on with ...


3

Good news is you are in right track and set up two equations correctly. However, I think the calculations would get easier if you have chosen your units of variables in $\pu{mol}$. Also, you may not needed to do more calculations since the answer is in $\pu{mol}$. Thus, your two equations should be: $$32x + 46y = 2.16 \;\text{.... (1), and}$$ $$\frac{3}{2}...


3

Case 2: Two Same Reactants $$2A + B \rightarrow C$$ Are the following correct? $$\frac{d[A]}{dt} = -2k \cdot [A][B]$$ $$\frac{d[B]}{dt} = -k \cdot [A][B]$$ $$\frac{d[C]}{dt} = 2k \cdot [A][B]$$ Let's call $k \cdot [A][B]$ the rate. Your equations simplify to: $$\frac{d[A]}{dt} = -2 \cdot \mathrm{rate}$$ $$\frac{d[B]}{dt} = -k \cdot \mathrm{rate}$$ ...


3

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off. If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving ...


2

There are several ways of determining stoichiometric formula from the known unit cell. Counting atoms [correctly] Perfectly covered in the answer by Curt F.; I would only like to propose to use data in a tabular form in order not to miss any of the atoms or improperly assign their environment. Briefly, not all the atoms you see on your image belong to the ...


2

A quick way to see what is going on without calculations is to move the origin of the unit cell a bit to the top, right, and back. This way, the atoms on the bottom face, on the left face and on the front face are no longer in the unit cell, and the eight atoms in the top-right-back corner are no longer shared by other unit cells. At the same time, because ...


2

You need to do a bit more math than that. In order to solve the majority of problems in chemistry you need to switch to the amounts of substances. So, let's use the following notations: $\omega$ – mass fraction; $m$ – mass; $n$ – amount; $M$ – molecular mass. I refer to $\ce{NaNO3}$ as compound $1$ and to $\ce{Na2SO4}$ as compound $2$. Let's solve this ...


2

First, let's solve the problem. Both relative atomic weight $A_\mathrm{r}$ and relative molecular weight $M_\mathrm{r}$ are a historical dimensionless terms denoting relative atomic mass and molecular mass, respectively: $$A_\mathrm{r} = \frac{m_\mathrm{a}}{m_\mathrm{u}}$$ $$M_\mathrm{r} = \frac{M_\mathrm{B}}{M_\mathrm{u}}$$ where $m$ corresponds to ...


2

Mass of $\ce{CuCl2.2H2O}$ needed can be found from its amount $n(\ce{CuCl2.2H2O})$ and molar mass $M(\ce{CuCl2.2H2O})$: $$m(\ce{CuCl2.2H2O}) = n(\ce{CuCl2.2H2O})\cdot M(\ce{CuCl2.2H2O}) \label{eqn:1}\tag{1}$$ The unknown amount of copper(II) chloride dihydrate $n(\ce{CuCl2.2H2O})$ can be found from the reaction stoichiometry: $$n(\ce{CuCl2.2H2O}) = \frac{...


2

When you say number of moles, it is a number to express the number of constituent species (atoms, in this case) in the given substance. It does not have any correlation to where that substance will be used and what will be its 'coefficient' in that reaction. When you say 10g Mg, you mean 0.411 moles of Mg (and vice versa) irrespective of whatever you are ...


2

The key point which you will study in advanced classes, is the concept of elementary reactions. When a given reaction is an elementary reaction, the stoichiometric coefficients become the power of the rate law or in better words, the reaction order corresponds to the stoichiometric coefficients. Definition of elementary reaction: An elementary reaction is a ...


2

N-factor Is the change of atom oxidation state multiplied by number of atoms changing this state. Therefore, n-factor of $\ce{I2}$ is $2$, n-factor of $\ce{Na2S2O3}$ is $1$. So, equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass, for molecular iodine it is half of its molar mass.


2

A balanced equation must have a mass balance i.e., the masses should be equal on both sides and charges must be balanced as well. Oxidation number is just a way of book-keeping. Nothing fundamental there. $$\ce{Fe^2+ + Cr2O7 + 14H+ -> 2 Cr^3+ + Fe^3+ 7 H2O}$$ What are you forgetting? It is mass balanced. Does $\ce{Cr2O7}$ exist? Hint: It is potassium ...


2

In acidic medium the equation must mention somewhere that some $\ce{H+}$ ions appear somewhere in the equation. In basic medium, the equation must mention somewhere that some $\ce{OH-}$ ions appear somewhere in the equation, For example, you may say that permanganate ion reacts in acidic conditions to produce $\ce{Mn^2+}$. In basic conditions, it could not ...


1

Only the chemical formulas of products and reactants of a chemical system are given. What mathematical methods can be used to determine all combinatorially possible chemical reactions between species of this reaction system? Let's take the example mentioned in the question, plus dihydrogen to make a point: $$\ce{CO2, C7H10N, H2O, O2, H2 and NO2}$$ We can ...


1

I agree with Poutnik's point that this is a poor question. Oxalic acid has two acidic protons so its equivalent weight is $\frac{1}{2}$ its molecular weight. Thus: Answer A is false. Answer B is true. Answer C is false since there are 2*100*0.2 = 40 milliequivalents of acid, but only 100*0.2 = 20 milliequivalents of base. Answer D requires some extra ...


1

Given the problem as stated, you did the problem correctly. However I would have done it a bit differently to make it easier to check. I like to do the problems in steps. I also dislike carrying a lot of fractions in intermediate calculations since I get confused easily. (This is basically M. Farooq's answer with some explanation.) Given $$\ce{2FeS2 + 11/...


1

Note the following relations from the equation you wrote: 1 mol $\ce{FeS2}$ = (11/4) mol $\ce{O2}$; 1 mol $\ce{FeS2}$ = (1/2) mol $\ce{Fe2O3}$; 1 mol $\ce{FeS2}$ = 2 mol $\ce{SO2}$. You already determined that $\ce{FeS2}$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $\ce{O2}$; This implies that 5 mol $\ce{FeS2}$ ...


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