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worty [1.4K]
3 years ago
10

Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: CO(g)+2H2(g)→CH3OH(g)CO(g)+2H2(g)→CH3O

H(g) A 1.55 LL reaction vessel, initially at 305 KK, contains carbon monoxide gas at a partial pressure of 232 mmHgmmHg and hydrogen gas at a partial pressure of 389 mmHgmmHg . Identify the limiting reactant and determine the theoretical yeild of methonal in grams.
Chemistry
1 answer:
zvonat [6]3 years ago
7 0

<u>Answer:</u> The limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT            ..........(1)

  • <u>For carbon monoxide:</u>

We are given:

P=232mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Putting values in equation 1, we get:

232mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{232\times 1.55}{62.3637\times 305}=0.0189mol

  • <u>For hydrogen gas:</u>

We are given:

P=389mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Putting values in equation 1, we get:

389mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{389\times 1.55}{62.3637\times 305}=0.0317mol

For the given chemical equation:

CO(g)+2H_2(g)\rightarrow CH_3OH(g)

By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of carbon monoxide

So, 0.0317 moles of hydrogen gas will react with = \frac{1}{2}\times 0.0317=0.01585mol of carbon monoxide

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

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

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 1 mole of methanol

So, 0.0317 moles of hydrogen gas will produce = \frac{1}{2}\times 0.0317=0.01585moles of methanol

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

Molar mass of methanol = 32 g/mol

Moles of methanol = 0.01585 moles

Putting values in equation 1, we get:

0.01585mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.01585mol\times 32g/mol)=0.507g

Hence, the limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams

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Answer:

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Explanation:

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7 0
3 years ago
A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled
marshall27 [118]

Answer:

0.257 L

Explanation:

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Final molarity of the solution =75.0 \mathrm{mM}

Final volume of the solution =?

From Law of Dilution, M_{f} V_{f}=M_{i} V_{i}

\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L

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7 0
3 years ago
The reaction of pyrrole with bromine forms predominantly __________. View Available Hint(s) The reaction of pyrrole with bromine
xenn [34]

Answer:

a) 2-bromopyrrole

Explanation:

Our options for this questions are:

a) 2-bromopyrrole

b) 2,3-dibromopyrrole

c) N-bromopyrrole

d) 3-bromopyrrole

To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.

The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the Br_2 and FeBr_3. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).

I hope it helps!

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Carla has 45$. Pineapples cost $5 each
ollegr [7]
She can buy 9 pineapples. I don't know why anyone would need that many pineapples
8 0
3 years ago
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If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield
Zigmanuir [339]
<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

<h2>Why?</h2>

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

ActualYield=37.6g\\TheoreticalYield=112.8g

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0
3 years ago
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