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Volgvan
3 years ago
9

Through stoichiometric calculations, the theoretical yield of indigo was calculated to be 1.70 g. However, when massed after the

experiment, only 0.73 g of indigo had been produced.
What is the percent yield of indigo?

A) 97.0%
B) 232.8%
C) 57.1%
D) 42.9%
Chemistry
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

                     %age Yield  =  42.9 %

Explanation:

Percentage Yield is defined as the actual yield divided by theoretical yield. It is given as,

                   %age Yield  =  Actual Yield / Theoretical Yield × 100 --- (1)

Also,

         Actual Yield is the amount of desired product which is obtained after an experiment is completed. We can also call it the real yield which is obtained practically. While, theoretical yield is the yield which is calculated on paper. Hence, we can say that the actual yield is always less than theoretical yield due to many reasons including reaction conditions and researcher handling problems.

Thus we are given following data;

                   Theoretical Yield  =  1.70 g

                   Actual Yield  =  0.73 g

Putting these values in Eq. 1,

                   %age Yield  =  0.73 g / 1.70 g × 100

                   %age Yield  =  42.9 %

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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

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CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

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No. of moles of H_2 = 2.40 mol

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                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

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8 0
4 years ago
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

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<u>Step 1:</u> Data given

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Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

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Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

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⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

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Step 4: Calculate heat

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Mass = 200 grams

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⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

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Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

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