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3 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

Chemistry

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

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Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago

g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con

Tanzania [10]

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

$t_{1/2}=\frac{0.693}{k}}$ .....(1)

Rate law expression for first order kinetics is given by the equation:

$t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}$ ......(2)

where,

k = rate constant

$t_{1/2}$ =Half life of the reaction = $2.42\times 10^3 s$

t = time taken for decay process = ?

$[A_o]$ = initial amount of the reactant = 0.163 M

[A] = amount left after time t = 66.8% of $[A_o]$

[A]= $\frac{66.8}{100}\times 0.163 M=0.108884 M$

$k=\frac{0.693}{2.42\times 10^3 s}$

$t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}$

t = 1,409.19 s

1 minute = 60 sec

$t=\frac{1,409.19 }{60} min=23.49 min$

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

6 0

3 years ago

Atoms form chemical bonds to satisfy the rule and to become .

soldi70 [24.7K]

Answer: Atoms form chemical bonds to satisfy the<u> Octet</u> rule and to become <u>stable.</u>

Explanation:

The tendency of atoms to attempt to get a noble gas configuration that is eight valence electrons is said to be octet rule. This is done to attain noble gas configuration and stability.

In order to attain stability the atoms tends to have eight electrons in its valence shell which can be obtained by either by sharing of electrons or complete transfer of electrons.

For example : As we know that the sodium has one valence electron, so if giving it up then the result in the same electron configuration as the neon and chlorine has seven valence electrons, so if it takes one it will have eight and the result in the same electronic configuration as the argon which is stable.

4 0

4 years ago

Identify the type of atom that generally forms covalent bonds

klasskru [66]

The answer is ionic atom

5 0

3 years ago

He resonance structures of carbon monoxide are shown below. show how each structure can be converted into the other using the cu

Ugo [173]

The two resonating structures of Carbon Monoxide are shown below.
The movement of electrons are shown by arrow.

Structure A:
In structure a A the formal charges of Carbon and Oxygen are zero. As Formal Charge is calculated as,

= # of valence electron - electrons in lone pairs + 1/2 bonding electrons electrons

For C:
= 4 - 2 + 4/2
= 4- 4
= 0

For O:
= 6 - 4 + 4/2
= 6- 6
= 0

Structure B:

Formal Charge on C:

= 4 - 2 + 6/2
= 4- 5
= -1

Formal Charge on O:

= 6 - 2 + 6/2
= 6 - 5
= +1

4 0

3 years ago