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Veronika [31]
4 years ago
12

Ron weasley isn't the best experimentalist. after an experiment he ends up with only 3.10 g of product. however, the same reacti

on has is a theoretical yield of 3.50 grams of product. what is the percent yield?
Chemistry
1 answer:
Veronika [31]4 years ago
4 0
The experiment should theoretically yield 3.50 g of the product.
However during the experiment it can yield only 3.10 g of the product.
actual yield is lesser than the theoretical yield, therefore we need to find the percentage yield.
Percentage yield can be calculated using the following equation 
Percent yield = \frac{actual  yield}{theoretical yield}  * 100%
Percent yield = 3.10 g / 3.50 g x 100% = 88.6%
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Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial pressure (P1) = 1atm

Initial temperature (T1) = 0°C = 0°C + 273 = 273K

Final temperature (T2) = 280°C = 280°C + 273 = 553K

Final pressure (P2) =...?

Step 2:

Determination of the new pressure of the gas.

Since the volume of the gas is constant, the following equation:

P1/T1 = P2/T2

will be used to obtain the pressure. This is illustrated below:

P1/T1 = P2/T2

1/273 = P2 / 553

Cross multiply

273x P2 = 553

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velikii [3]

Answer:

A. ΔH∘rxn =  21.9 KJ/mol

B. ΔH∘rxn = 103 KJ/mol

C. C2H5OH + 3O2 → 2CO2 + 3H2O

Explanation:

A.

The standard reaction equation is given as:

aA + bB → cC + dD

Its standard enthalpy is given as:

ΔH∘rxn = cΔH∘f(C) + dΔH∘f(D) − aΔH∘f(A) − bΔH∘f(B)

Reaction given to us is:

H2O(l) + CCl4(l) → COCl2(g) + 2HCl(g)

So, its standard enthalpy will be:

ΔH∘rxn = (1)ΔH∘f(CoCl2 (g)) + (2)ΔH∘f(HCl(g)) − (1)ΔH∘f(H2O(l)) − (1)ΔH∘f(CCl4(l))

using the values from table:

ΔH∘rxn = - 218.8 KJ/mol + (2)(- 92.3 KJ/mol) - (- 285.8 KJ/mol) - (- 139.5 KJ/mol)

<u>ΔH∘rxn =  21.9 KJ/mol</u>

<u></u>

B.

Reaction given to us is:

2A + B ⇌ 2C + 2D

So, its standard enthalpy will be:

ΔH∘rxn = (2)ΔH∘f(C) + (2)ΔH∘f(D) − (2)ΔH∘f(A) − (1)ΔH∘f(B)

using the values from table:

ΔH∘rxn = (2)181 KJ/mol + (2)(- 523 KJ/mol) - (2)(- 225 KJ/mol) - (- 337 KJ/mol)

<u>ΔH∘rxn = 103 KJ/mol</u>

<u></u>

C.

Balanced equation for combustion of ethanol is:

<u>C2H5OH + 3O2 → 2CO2 + 3H2O</u>

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