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3 years ago

How many grams of C2H6 are needed to react with 54 grams of O2?

Chemistry

1 answer:

Dmitrij [34]3 years ago

4 0

Answer:

Explanation:

in a combustion of ethane 2 moles of ethane react with 7 moles of O2

now no of moles in 54 gram of O2=mass/ molar mass

moles =54/32=1.7 moles

if 7 moles of O2 required 2 moles of ethane then 1.7 mole required=?

7 moles of O2=2 moles of C2H6

1.7 moles of O2=1.7*2/7=0.5 moles of C2H6

0.5 moles of C2H6 contain how much grams=?

mass= moles*molar mass=0.5*30=15

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After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage

Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have 2 reactions, the formation of the grignard reagent and the formation of the alcohol. The first step then is the calculation of the maximum amount of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the following conversion ratios:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide ( $C_6H_5Br$ )= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and $C_6H_5Br$ = 1:1

$2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg$

$9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br$

The smallest value is the mol of bromobenzene therefore 0.0402 mol of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of formation of the alcohol. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

$4.83~g~\frac{1~mol}{136.14~g}=0.35~mol$

The we have to divide by the coefficient of each reactive in the balance reaction. So:

$\frac{0.35~mol}{1}=0.35$

$\frac{0.0402~mol}{2}=0.0201$

Therefore the limiting reagent would be the phenylmagnesium bromide. Now, the molar ratio between the phenylmagnesium bromide and triphenyl carbinol is 2:1, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to grams of triphenyl carbinol:

Molar mass of triphenyl carbinol= 260.33 g/mol

$0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol$

Finally, we have to divide the obtanied solid by the calculated one:

$Percentage=\frac{5}{5.23}*100=95.6\%$

5 0

3 years ago

Is The reaction time is different for different stimuli true or false?

MArishka [77]

Answer: true

Explanation: In this example, the processes (perceive, process, and respond), are done in a matter of milliseconds, but reaction time can vary depending on a variety of factors: Complexity of the stimulus-The more complex the stimulus, the more information that has to be processed, the longer this process will take.

5 0

3 years ago

A group of students is investigating how the addition of salt impacts the floatation of an egg in water. Four identical cups wer

soldier1979 [14.2K]

Answer: The correct answer is Option 1.

Explanation:

Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.

We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.

Hence, the correct answer is Option 1.

4 0

3 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

3 years ago

When potassium metal is placed in water, a large amount of energy is released as potassium hydroxide and hydrogen gas are produc

blsea [12.9K]

Answer: The given reaction redox reaction not a combustion reaction.

Explanation:

Redox Reaction is defined chemical reaction in which oxidation and reduction takes place simultaneously. In Oxidation addition of oxygen occurs and in reduction addition of hydrogen occurs.

$2K+2H_2O\rightarrow 2KOH+H_2$

Potassium is getting oxidized by addition of oxygen.

Water us reduced to hydrogen gas by removal of oxygen

Combustion reaction is defined as chemical reaction in which burning of a chemical compound or element takes place in presence of oxygen occurs.

In the given reaction, oxygen molecule is absent which means this reaction is not combustion reaction.

Hence ,the given reaction redox reaction not a combustion reaction.

8 0

3 years ago