0.1 Analyzing chemical equations  (Page 3/3)

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Limiting reactant

The participating mass of the reactants may not be exactly same as required by balanced chemical reaction. The reactant which limits the progress of reaction is called limiting reactant (reagent). If gA and gB be the mass of two reactants, then moles present are :

${n}_{A}=\frac{{g}_{A}}{{M}_{A}}$

${n}_{B}=\frac{{g}_{B}}{{M}_{B}}$

According to mole concept :

$\text{x moles of A}\equiv \text{y moles of B}$

$\text{nA moles of A}\equiv \frac{y{n}_{A}}{x}\text{moles of B}$

For reaction to complete as the limiting case,

$\frac{y{n}_{A}}{x}={n}_{B}$

$y{n}_{A}=x{n}_{B}$

For A to be limiting,

${n}_{B}>\frac{y{n}_{A}}{x}$

$y{n}_{A}

For B to be limiting,

${n}_{B}<\frac{y{n}_{A}}{x}$

$x{n}_{B}

$y{n}_{A}>x{n}_{B}$

Problem : Equal amounts of iron and sulphur are heated together to form FeS. What fraction of iron or Sulphur are converted into FeS. Here, ${A}_{Fe}=56$ and ${A}_{S}=32$ .

Solution : The chemical reaction involved in the conversion is :

$Fe+S\to FeS$

$\text{1 mole of Fe}\equiv \text{1 moles of S}$

Here, x=y=1. Now, equal amounts are used. Let “g” gm of each takes part in the reaction. The numbers of moles corresponding to “g” gm are :

${n}_{Fe}=\frac{g}{56}$

and

$y{n}_{Fe}=\frac{g}{56}$

Similarly,

$x{n}_{S}=\frac{g}{32}$

Here, $y{n}_{Fe} . It means that iron is in short supply and is the limiting reactant. Clearly, all iron is converted into FeS. Hence, fraction of iron used is 1. On the other hand, the moles of Sulphur converted are $\frac{x}{56}$ only as reaction is limited by iron.

$⇒\text{Mass of Sulphur converted in the reaction}=\frac{x}{56}X{M}_{S}=\frac{32x}{56}$

$⇒\text{Fraction of suphur converted}=32x/56x=\frac{32}{56}=0.571$

Multiple reactions

Some chemical reaction under analysis involves simultaneous or concurrent reactions. Consider the example in which a mixture of KCl and KI are treated with excess of silver nitrate ( $AgN{O}_{3}$ ) :

$KCl+AgN{O}_{3}\to KN{O}_{3}+AgCl$

$KI+AgN{O}_{3}\to KN{O}_{3}+AgI$

We apply mole concept to two concurrent reactions as :

$\text{1 mole of KCl}\equiv \text{1 mole of}AgN{O}_{3}\equiv \text{1 mole of}KN{O}_{3}\equiv \text{1 mole of AgCl}$

and

$\text{1 mole of KI}\equiv \text{1 mole of}AgN{O}_{3}\equiv \text{1 mole of}KN{O}_{3}\equiv \text{1 mole of AgI}$

Problem : 2.2 g of a mixture of KCl and KI yields 3.8 gm of AgCl and AgI, when treated with excess silver nitrate ( $AgN{O}_{3}$ ). Find the mass of KCl and KI in the original mixture.

Solution : Since the mixture is treated with excess silver nitrate, both chloride and iodide are completely consumed, forming silver chloride and iodide. Let the initial mixture contains x and y grams of KCl and KI respectively. Then, according to question :

$x+y=2.2$

This is first of two linear equations, which will be used to determine x and y. Now, chemical reactions are :

$KCl+AgN{O}_{3}\to KN{O}_{3}+AgCl$

$KI+AgN{O}_{3}\to KN{O}_{3}+AgI$

The amount of AgCl and AgI produced is 3.8 gm. Applying mole concept to first equation,

$\text{1 mole of KCl}\equiv \text{1 mole of AgCl}$

$\frac{x}{\left(39+35.5\right)}\text{moles of KCl}\equiv \frac{x}{\left(39+35.5\right)}\text{moles of AgCl}\equiv \frac{x}{74.5}\text{moles of AgCl}$

Therefore, mass of AgCl is :

$\frac{x}{74.5}\text{moles of AgCl}=\frac{x}{74.5}X\left(108+35.5\right)\phantom{\rule{1em}{0ex}}gm=\frac{143.5x}{74.5}\phantom{\rule{1em}{0ex}}gm$

Similarly, applying mole concept to second equation,

$\text{1 mole of KI}\equiv \text{1 mole of AgI}$

$\frac{x}{\left(39+127\right)}\text{moles of KCl}\equiv \frac{y}{\left(39+127\right)}\text{moles of AgI}\equiv \frac{y}{166}\text{moles of AgI}$

Therefore, mass of AgI is :

$\frac{y}{166}\text{moles of AgI}=\frac{y}{166}X\left(108+127\right)\phantom{\rule{1em}{0ex}}gm=235\frac{y}{166}\phantom{\rule{1em}{0ex}}gm$

According to question,

$\frac{143.5x}{74.5}+\frac{235y}{166}=3.8$

In order to render coefficient of first term of the equation equal to 1 and hence simplify the equation, we multiply the equation by $\frac{74.5}{143.5}$ ,

$x+\frac{74.5X235y}{143.5X166}=\frac{3.8X74.5}{143.5}$

$⇒x+0.73y=1.97$

Thus, we have two linear equations and two unknowns. Substituting for y in terms of x from first linear equation derived earlier, we have :

$⇒x+0.73\left(2.2-x\right)=1.97$

$⇒x+1.61-0.73x=1.97$

$⇒0.27x=0.36$

$⇒x=1.34\phantom{\rule{1em}{0ex}}gm$

$⇒y=2.2-x=2.2-1.34=0.86\phantom{\rule{1em}{0ex}}gm$

Chain reaction

A chemical process involves multiple reactions in which product of one reaction becomes reactant in the second reaction. Consider example of formation of sulphuric acid from pyrite( $Fe{S}_{2}$ ) :

$4Fe{S}_{2}+11{O}_{2}\to 2F{e}_{2}{O}_{3}+8S{O}_{2}$

$2S{O}_{2}+{O}_{2}\to 2S{O}_{3}$

$S{O}_{3}+{H}_{2}O\to {H}_{2}S{O}_{4}$

Here, first reaction supplies $S{O}_{2}$ required as reactant in the second reaction. We, therefore, need to multiply second reaction 4 times so that $S{O}_{2}$ molecules are balanced. Similarly, third reaction should be multiplied by 8 to match $8S{O}_{3}$ molecules produced in the second reaction. Thus, balanced chain reaction is :

$4Fe{S}_{2}+11{O}_{2}\to 2F{e}_{2}{O}_{3}+8S{O}_{2}$

$8S{O}_{2}+4{O}_{2}\to 8S{O}_{3}$

$8S{O}_{3}+8{H}_{2}O\to 8{H}_{2}S{O}_{4}$

From balanced chain equations, we conclude that :

$⇒\text{4 moles of}\phantom{\rule{1em}{0ex}}Fe{S}_{2}\equiv \text{8 moles of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

$⇒\text{1 mole of}\phantom{\rule{1em}{0ex}}Fe{S}_{2}\equiv \text{2 moles of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

Alternatively, we can argue that one molecule of $Fe{S}_{2}$ contains 2 atoms of Sulphur(S). On the other hand, one molecule of ${H}_{2}S{O}_{4}$ contains 2 atoms of Sulphur(S). Further, there is no loss or gain of Sulphur in chain reactions. Hence, to conserve mass of suphur,

$⇒\text{1 mole of}\phantom{\rule{1em}{0ex}}Fe{S}_{2}\equiv \text{2 moles of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

Problem : How many grams of sulphuric acid can be obtained from 100 gm of pyrite?

Solution : Here,

$⇒\text{1 mole of}\phantom{\rule{1em}{0ex}}Fe{S}_{2}\equiv \text{2 moles of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

$⇒\text{(56+2X32) gm of}\phantom{\rule{1em}{0ex}}Fe{S}_{2}\equiv \text{2X(2X1+32+4X16) moles of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

$⇒\text{120 gm of}\phantom{\rule{1em}{0ex}}Fe{S}_{2}\equiv \text{196 gm of}{H}_{2}S{O}_{4}$

Applying unitary method,

$⇒\text{mass of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}=\frac{192}{120}X100=160\phantom{\rule{1em}{0ex}}gm$

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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Sherica
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Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
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yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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what is system testing?
AMJAD
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Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Stoichiometry. OpenStax CNX. Jul 05, 2008 Download for free at http://cnx.org/content/col10540/1.7
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