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Leno4ka [110]
2 years ago
15

When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?

Chemistry
1 answer:
Afina-wow [57]2 years ago
3 0

Answer:

Y=75.6\%

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2}  *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%

Best regards.

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The town of Natrium, West Virginia, derives its name from the sodium produced in the electrolysis of molten sodium chloride (NaC
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Explanation:

The reaction equation will be as follows.

           Na^{+} + e^{-} \rightarrow Na(s)

Hence, moles of Na = moles of electron used

Therefore, calculate the number of moles of sodium as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{4500 g}{23 g/mol}    (as 1 kg = 1000 g)

                             = 195.65 mol

As,     Q = n \times F       where F = Faraday's constant

              = 195.65 mol \times 96500 C

              = 1.88 \times 10^{7} mol C

Relation between electrical energy and Q is as follows.

               E = Q \times V

Hence, putting the given values into the above formula and then calculate the value of electricity as follows.

              E = Q \times V

                 = 1.88 \times 10^{7} \times 5

                 = 9.4 \times 10^{7} J

As 1 J = 2.77 \times 10^{-7} kWh

Hence,      \frac{9.4 \times 10^{7}}{2.77 \times 10^{-7}} kWh

                = 3.39 kWh

Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.

7 0
2 years ago
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Inga [223]
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1 year ago
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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
zaharov [31]

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

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7 0
2 years ago
What is the hydronium ion concentration of a solution whose pH is 7.30
Assoli18 [71]
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[ H₃O⁺ ] = 10 ^ - 7.30

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hope this helps!
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1 year ago
1. Which liquid sample is a pure substance?
IRISSAK [1]

The whole Activity , poem and paragraph is missing in the question.

Answer:

(1) Liquid A

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Explanation:

Using this part of the given poem

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Solid A is melting completely so Solid A is a pure substance.

6 0
1 year ago
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