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3 years ago

7

Scientists are often interested in knowing the molar heat of combustion the heat released during the combustion of one molar of

a substance use a periodic table to find molar masses.
How many moles of ethanol are present in the sample?

? Mol

2 answers:

Butoxors [25]3 years ago

6 0

Answer:

.004993

Explanation:

lozanna [386]3 years ago

6 0

Answer: .004993

Explanation:

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In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde, CH3CHO, containing manganese(II) acetate (catal

grigory [225]

<u>Answer:</u> The mass of acetic acid that can be produced is 30.24 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For acetaldehyde:</u>

Given mass of acetaldehyde = 22.2 g

Molar mass of acetaldehyde = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetaldehyde}=\frac{22.2g}{44g/mol}=0.504mol$

<u>For oxygen gas:</u>

Given mass of oxygen gas = 12.6 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{12.6g}{32g/mol}=0.394mol$

The given chemical equation follows:

$2CH_3CHO(l)+O_2(g)\rightarrow 2CH_3COOH(l)$

By Stoichiometry of the reaction:

2 moles of acetaldehyde reacts with 1 mole of oxygen gas

So, 0.504 moles of acetaldehyde will react with = $\frac{1}{2}\times 0.504=0.252mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, acetaldehyde is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of acetaldehyde produces 2 moles of acetic acid

So, 0.504 moles of acetaldehyde will produce = $\frac{2}{2}\times 0.504=0.504moles$ of acetic acid

Now, calculating the mass of acetic acid from equation 1, we get:

Molar mass of acetic acid = 60 g/mol

Moles of acetic acid = 0.504 moles

Putting values in equation 1, we get:

$0.504mol=\frac{\text{Mass of acetic acid}}{60g/mol}\\\\\text{Mass of acetic acid}=(0.504mol\times 60g/mol)=30.24g$

Hence, the mass of acetic acid that can be produced is 30.24 grams

8 0

3 years ago

Identify the type of reaction: 2AgCl + BaBr2 -> 2AgBr + BaCl2

Marta_Voda [28]

Answer:

C

Explanation:

Just look at the reactants.

2AgCl + BaBr2

The first reactant is made of 2 elements.

The second reactant is made of 2 elements.

It can't be a decomposition. At this level there is only one reactant made of 2 elements. Something like

2MgO ===> 2Mg + O2

is a decomposition. One compound breaking down into 2 elements Mg and O.

It can't be a combustion. One of the reactants in a combustion is oxygen. Those equations look like

C3H8 + 5O2 ==> 3CO2 + 4H2O

That would be what a combustion looks like

It can't be a single replacement. They look like

Mg + CuO ===> Cu + MgO

There are elements on both sides of the reaction.

that leaves a double replacement which I wrote about how you distinguish it.

6 0

2 years ago

Consider the chemical system co + cl2 cocl2; k = 4.6 ? 109 l/mol. 2. how do the equilibrium concentrations of the reactants comp

ikadub [295]

<span>they are much bigger because </span><span>equilibrium constant shows ratio of </span><span>concentrations of the reactants to the equilibrium concentration of the product.</span>

3 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

<u>Answer:</u> The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

<u>Explanation:</u>

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago

Why is it logical to assume that the hydrogen ion concentration in an aqueous solution of a strong monoprotic acid equals the mo

Masja [62]

Strong acids are those that dissociate completely into their ions. So the dissociation equation for a strong acid is:
HA → H⁺ + A⁻

It is visible from the equation that the number of moles of hydrogen ions released is equivalent to the number of moles of acid. For a given volume,
[HA] → [H]⁺ + [A]⁻
Thus, the assumption is logical and fairly accurate

6 0

2 years ago