1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rom4ik [11]
3 years ago
11

A reaction starts with 20.0 g of lithium hydroxide (LiOH) and actually produces 31.0 g of lithium chloride (LiCl), what is the p

ercent yield? (Hint: First calculate the theoretical yield of lithium chloride (LiCl))
64.5%
88.6%
81.5%
92.8%

Chemistry
2 answers:
Kisachek [45]3 years ago
5 0

Answer:

88.6%

Explanation:

Hello,

In this case, considering the given reaction, we notice a 1:1 molar relationship between lithium hydroxide (molar mass=23.95 g/mol) and lithium chloride (molar mass=42.394 g/mol), for that reason, we are able to compute the theoretical yield of lithium chloride by stoichiometry:

m_{LiCl}^{theoretical}=20.0gLiOH*\frac{1molLiOH}{23.95gLiOH}*\frac{1molLiCl}{1molLiOH} *\frac{42.394 gLiCl}{1molLiCl}=35.4gLiCl

Next, by knowing the actual yield of 31.0 g, we compute the percent yield as:

Y=\frac{m_{LiCl}^{actual}}{m_{LiCl}^{theoretical}} *100\%=\frac{31.0g}{35.4g}*100\%\\ \\Y=87.6\%

Therefore, among the given, the answer should be 88.6%

Best regards.

podryga [215]3 years ago
3 0

Answer:

87.6 %

Explanation:

Step 1: Write the balanced equation

LiOH + KCl ⇒ LiCl + KOH

Step 2: Calculate the theoretical yield of LiCl

We will use the following relations:

  • The molar mass of LiOH is 23.95 g/mol.
  • The molar ratio of LiOH to LiCl is 1:1.
  • The molar mass of LiCl is 42.39 g/mol.

The theoretical yield of LiCl from 20.0 g of LiOH is:

20.0gLiOH \times \frac{1molLiOH}{23.95gLiOH} \times \frac{1molLiCl}{1molLiOH} \times \frac{42.39gLiCl}{1molLiCl} = 35.4gLiCl

Step 3: Calculate the percent yield of LiCl.

We will use the following expression.

\%yield = \frac{real\ yield}{theoretical\ yield} \times 100\% = \frac{31.0g}{35.4g} \times 100\% = 87.6 \%

You might be interested in
Is respiration like burning?Explain your answer.​
SIZIF [17.4K]

Answer:

Yes

Explanation:

The process of respiration and burning are similar in the following ways: Both respiration and burning require oxygen. Energy is released during both respiration and burning. The products produced (carbon dioxide and water) are the same for both respiration and burning.

3 0
2 years ago
Read 2 more answers
Terbium-147 undergoes positron emission to become a stable atom. What is that stable atom?
ivanzaharov [21]

We have to get the stable atom formed after positron emission from Terbium-147.

The stable atom is (D) ₆₄Gd¹⁴⁷.

Positron is radioactive decay. Positron is a type of beta particle β⁺.

Positron emission decreases proton number relative to neutron number, positron decay results in nuclear transmutation, changing an atom of one chemical element with an atomic number that is less by one.

Terbium on positron emission produces Gadolinium with one atomic number less than Terbium. So, the positron emission reaction is as shown below:

Tb¹⁴⁷→ ₆₄Gd¹⁴⁷ + ₁e⁰

7 0
3 years ago
Paraphrase the first part of newton’s first law of motion
Zarrin [17]
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction UNLESS acted upon by an unbalanced force.
5 0
3 years ago
Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several di
Ira Lisetskai [31]

Answer : The correct rate law for the reaction is,

\text{Rate}=k[CHCl_3][Cl_2]^{1/2}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)

Rate law expression for the reaction:

\text{Rate}=k[CHCl_3]^a[Cl_2]^b

where,

a = order with respect to CHCl_3

b = order with respect to Cl_2

Expression for rate law for first observation:

0.0035=k(0.010)^a(0.010)^b ....(1)

Expression for rate law for second observation:

0.0069=k(0.020)^a(0.010)^b ....(2)

Expression for rate law for third observation:

0.0098=k(0.020)^a(0.020)^b ....(3)

Expression for rate law for fourth observation:

0.027=k(0.040)^a(0.040)^b ....(4)

Dividing 1 from 2, we get:

\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1

Dividing 2 from 3, we get:

\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}

Calculation used :

1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}

Thus, the rate law becomes:

\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}

7 0
3 years ago
If an object reflects green and absorbs all other colors, it will appear _____?
nataly862011 [7]
It will appear to be the color red
7 0
3 years ago
Other questions:
  • When is the small size of gas particles taken into account?
    9·1 answer
  • the point on earths surface directly above where an earthquake happen is called? a.)focus b.)epicenter c.)hypocenter d.)locus​
    12·2 answers
  • A solution is made by dissolving
    9·1 answer
  • Which describes the composition of matter in the figure C?
    5·1 answer
  • Maria examines a mixture that appears to be the same throughout. However, when she looks at a sample using a magnifying lens, sh
    11·2 answers
  • A compound is 39.97% carbon, 13.41% hydrogen, and 46.62% nitrogen. What is the empirical formula?
    15·2 answers
  • Genetic material (_____) is passed from cell to cell.
    13·1 answer
  • How many moles of C6H12O6 does 8.2 x 1023 molecules represent
    6·2 answers
  • What is the mass in grams of 0.5440 moles of sodium?
    15·1 answer
  • 21.10g of NaOH and Ba3(OH)2 mixture is dissolved water to prepare 1.0dm³ Solution. To neutralize 25.OO mL of this solution needs
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!