answer.
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
Brrunno [24]
2 years ago
9

A 1.225 g sample mixture of lithium hydrogen carbonate is decomposed by heating to produce 0.660 g lithium carbonate. Calculate

the theoretical yield and percent yield of Li2CO3
Chemistry
1 answer:
Eva8 [605]2 years ago
4 0

Answer:

  • The theoretical yield of lithium carbonate = 0.6659 g.
  • The percent yield = 99.11%.

Explanation:

  • Lithium hydrogen carbonate is decomposed by heating according to:

<em>2LiHCO₃(s) → Li₂CO₃(s) + H₂CO₃(g).</em>

<em>It is clear that 2.0 moles of lithium hydrogen carbonate produce 1.0 mole of lithium carbonate.</em>

<em />

  • We need to calculate the no. of moles of (1.225 g) lithium hydrogen carbonate decomposed:

no. of moles of lithium hydrogen carbonate = mass/molar mass = (1.225 g)/(67.96 g/mol) = 0.018 mol.

  • 0.02 mol of lithium hydrogen carbonate is decomposed to produce (0.02 mo /2 = 0.009 mol) of lithium carbonate.

<em>∴ The theoretical yield of lithium carbonate = no. of moles x molar mass</em> = (0.009 mol)(73.89 g/mol) = <em>0.6659 g.</em>

<em>∵ The percent yield = (actual yield/theoretical yield) x 100 = (0.660 g / 0.6659 g) x 100 = 99.11%.</em>

You might be interested in
What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio
Veseljchak [2.6K]

Answer:

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol

Mass of methane burnt = 530 g

Moles of methane burnt = \frac{530 g}{16 g/mol}=33.125 mol

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ

-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

7 0
2 years ago
In the chemical reaction NaHCO3 + CH3COOH → CH3COONa + H2O +CO2, 83 g of sodium bicarbonate reacts with 70 g of acetic acid. Whi
krok68 [10]
MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........

84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH

We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.

:•)
3 0
2 years ago
Suppose a layer of oil is on the top of a beaker of water. Water in many oils is slightly soluble, so its concentration is so lo
rodikova [14]

Explanation:

It is given that energy to transfer one water molecule is 2.208 \times 10^{-20} J/molecule

As it is known that in 1 mole there are 6.022 \times 10^{23} atoms.

So, energy in 1 mole = 2.208 \times 10^{-20} \times 6.022 \times 10^{23} J/mol

                                  = 13.3 kJ/mol

As,    log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}

Putting the given values in the above formula as follows.

                log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}

               log \frac{k_{2}}{k_{1}} = \frac{13.3 kJ/mol}{2.303 \times 8.314 atm L/mol K} \times \frac{293 K - 283 K}{283K \times 293 K}

                  log \frac{k_{2}}{k_{1}} = 0.08377

                       \frac{k_{2}}{k_{1}} = 1.213 = \frac{Concentration_{2}}{Concentration_{1}}

                 Concentration_{2} = 1.213 \times Concentration_{1}

                                             = 1.213 \times 9 \times 10^{-4}

                                             = 10.915 \times 10^{-4} water molecules per oil molecule

Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is 10.915 \times 10^{-4}.

             

4 0
2 years ago
Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a __________
Vesna [10]
Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a ____________ mixture. A) heterogeneous
7 0
2 years ago
Read 2 more answers
When 0.270 mol of a nondissociating solute is dissolved in 410.0 mL of CS2, the solution boils at 47.52 ∘C. What is the molal bo
grandymaker [24]

Answer:

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the <em>boiling-point elevation formula</em>:

  • <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.

So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.

molality = mol solute / kg solvent

  • Density of CS₂ = 1.26 g/cm³
  • Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg

molality = 0.270 mol / 0.5166 kg = 0.5226 m

Now we <u>solve for Kb</u>:

<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

  • 47.52 °C - 46.3 °C = Kb * 0.5226 m
  • Kb = 0.428 m/°C
3 0
2 years ago
Other questions:
  • What will be the pressure in atmospheres in a 750.0 mL vessel containing 5.00 g of argon gas at 15°C?
    15·1 answer
  • You are on a field trip to a nearby lake for biology class and want to perform a quick analysis of the water’s approximate pH le
    6·2 answers
  • Write a net ionic equation for the reaction that occurs when nickel(ii carbonate and excess hydrobromic acid (aq are combined.
    14·1 answer
  • Twenty-seven milliliters of an acid with an unknown concentration are titrated with a base that has a concentration of 0.55 M. T
    10·2 answers
  • A compound that contains both potassium and oxygen formed when potassium metal was burned in oxygen gas. the mass of the compoun
    7·2 answers
  • QUESTION 18 Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) 3O 2 (g) 2Al 2O
    15·1 answer
  • A scientist discovers a deep bowl-like divot under the ocean off the coast of eastern Mexico that is many kilometers across. The
    10·2 answers
  • What is the difference between matter and non matter ?
    7·2 answers
  • This element has three fewer protons than oganesson 
    8·1 answer
  • Calculate the molar mass of a 2.89 g gas at 346 ml, a temperature of 28.3 degrees Celsius, and a pressure of 760 mmHg.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!