CHEMISTRY COLLEGE

Scaled Synthesis of Alum. Show your calculations for:a.the experimental scaling factor giving rise to a 15.0 g theoretical yield;b.the corrected volumes of KOH and H2SO4; andc.the theoretical yield of alum based on the actual amount of Al used.Make sure you carefully show each step for these calculations.

Answers

Answer 1

Answer:

Answer:

Explanation:

You don't give your experimental data, so I shall assume:

Mass of Al = 1.07 g

20 mL of 3 mol·L⁻¹ KOH

20 mL of 9 mol·L⁻¹ H₂SO₄

The overall equation for the reaction is

Mᵣ: 26.98 474.39

2Al + 2KOH +4H₂SO₄ + 22H₂O ⟶ 2K[Al(SO₄)₂]·12H₂O + 3H₂

m/g: 1.07

(c) Theoretical yield of alum

(i) Moles of Al

$\text{Moles of Al} = \text{1.07 g Al} * \frac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.039 66 mol Al}$

(ii) Moles of alum

$\text{Moles of alum} = \text{0.039 66 mol Al} * \frac{\text{2 mol alum }}{\text{2 mol Al}} = \text{0.039 66 mol alum \n}$

(iii) Theoretical yield of alum

$\text{Mass of alum} = \text{0.039 66 mol alum} * \frac{\text{474.39 g alum}}{\text{1 mol alum}} = \textbf{18.8 g alum}$

(a) Scaling factor for 15.0 g alum

You want a theoretical yield of 15.0 g, so you must scale down the reaction.

$\text{Scale factor} = (15.0)/(18.8) = \mathbf{0.798}$

(b) Corrected volumes of NaOH and H₂SO₄

V = 0.798 × 20 mL = 16 mL

Related Questions

Energy travels in what direction?
Calculate the percent saturated fat in the total fat in butter
Write the numbers in scientific notation.291.7 = 2.917X10%where x =0.0960 -X10%where x =
Models can have the samegeneral appearance as real-ufeobjects.True or false
When radioactive uranium decays to produce thorium, it also emits a particle. As seen in the balanced nuclear equation, this particle can BEST be described asA) a helium atom. B) an alpha particle or a helium atom. C) a beta particle or a hydrogen nucleus. D) an alpha particle or a helium nucleus.

Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

Learn more about mole fractions here:

brainly.com/question/29808190

#SPJ3

Select ALL factors in conservation: *Social conditions such as need for electricity, famine, and war.
Scientific data related to the ecosystem and the effect of environmental changes.
Political action by governments and other organizations such as environmental
protection groups.
Economic issues such as cost of wood products, fuel for heat, price of electricity, and
income levels of local people.

Answers

Answer:

ALL of these are factors in conservation.

Explanation:

What is anology?
What is microscope?

Answers

Microphone thing is a little bit better but it was ok I gotta it looked good but it didn’t get me wrong it I made a comment on my phone number so ttyttttytt was the day I wanna was a good day and then I’ll send ya my stuff and I get to see ya my mom stuff is the best thing ever I just wanna was a

Counting atoms worksheet

Answers

Answer: if you are looking for a worksheet go to teachers pay s teachers or Pinterest

Explanation: these are very reliable sources to find good worksheets

Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)? A. NO (aq)
B. 2Ba (aq)
C. 2Ag (aq)
D. CI(aq)

Answers

Answer:

Option C

Explanation:

Consider the ionic equation of this chemical equation. We are given barium chloride and silver nitrate as the reactants, and silver chloride and barium nitrate as the products. We can thus conclude that the ionic equation ( not balanced yet ) should be as follows -

Ba( 2 + ) + Cl ( - ) + Ag ( + ) + NO3 ( - ) ------> AgCl + Ba( 2 + ) + NO3( - )

As you can see these compounds are present in aqueous solutions, and are thus dissociated.

______________________________________________________

Now let us take a look at the number of elements on the reactant and product sides, and balance this chemical equation out -

Ba( 2 + ) + 2Cl ( - ) + 2Ag ( + ) + 2NO3 ( - ) ------> 2AgCl + Ba( 2 + ) + 2NO3( - )

Solution = Option C!

Draw the Lewis structure for XeCl2 and answer the following questions.How many valence electrons are present in this compound?
How many bonding electrons are present in this compound?
How many lone pair (non-bonding) electrons are present in this compound?

Answers

Answer:

Valence electrons in XeCl2 = 8 + 7 + 7 = 22.

Bonding electrons = 4.

Nonbonding electrons = 18.

Explanation:

Hello.

In this case, you can see the Lewis structure on the attached picture, in which you can see that there are since xenon has 8 valance electrons and each chlorine has 7 valence electrons, the total amount of valence electrons is:

Valence electrons in XeCl2 = 8 + 7 + 7 = 22.

Moreover, since each chlorine atom is bonding with one of the eight electrons of xenon (Lewis structure), we can see there are 4 bonding electrons.

Finally, since there are six nonbonding electrons per chlorine atom and six nonbonding electrons in xenon, the overall nonbonding electrons are:

Nonbonding electrons in XeCl2 = 6 + 6 + 6 = 18.

Regards.

Other Questions

Can aniline be nitrated directly?

In metallic bonds, the mobile electrons surrounding the positive ions are called a(n)

Research Hypatia's achievements in the worldof science. What is she most known for?Write down three interesting facts about herlife.