2KClO3 —> 2KCl2+ 3O2

For the reaction of reducing benzil (MW 210.23 g/mol) with sodium borohydride (MW 37.83 g/mol), if 2.56 g of benzil and 0.38 g o

disa [49]

Answer:

NaBH₄

Explanation:

First, we need to write the chemical formula of all the compounds:

Benzil: C₁₄H₁₀O₂

Sodium Borhydride: C₁₄H₁₀O₂

Hydrobenzoin: C₁₄H₁₄O₂

Now, let's write the reaction that is taking place and write all the products:

C₁₄H₁₀O₂ + 2NaBH₄ + 2H₂O -----------> C₁₄H₁₄O₂ + 2BH₃ + 2NaOH

We can see that the reaction is already balanced, so we don't need to do anything else.

The question of this exercise is to determine the limiting reagent of the reaction, in other words, the reagent that controls the reaction and produces the 2.22 g of the hydrobenzoin. And to know this we need to see the mole ratio in both reactants, and compare them to the given moles (That can be obtained with the given masses and MW)

According to the above reaction, we have a mole ratio of 1:2, so, let's calculate the moles of benzil and the borohydride, and see which of them is the limiting reactant:

moles C₁₄H₁₀O₂ = 2.56 / 210.23 = 0.0122 moles

moles NaBH₄ = 0.38 / 37.83 = 0.01 moles

moles C₁₄H₁₄O₂ = 2.22 / 214.26 = 0.0103 moles

We have the moles of every species, now, let's see the mole ratio

If 1 mole of C₁₄H₁₀O₂ -----------> 2 moles of NaBH₄

Then 0.0122 moles C₁₄H₁₀O₂ ----------> X moles of NaBH₄

Solving for X:

X = 0.0122 * 2 / 1 = 0.0244 moles of NaBH₄ are required.

However, we only have 0.01 moles of NaBH₄, and we need so much more of this to completely react with the moles of the benzil. Therefore we can safely assume that the limiting reagent is the NaBH₄

Another data that we can use for this, is the fact the produced moles were 0.0103, and this value is nearest to the moles of NaBH₄ rather than the moles of the benzil.

<h2>So, in conclusion, Limiting reagent NaBH₄</h2>

Hope this helps

7 0

3 years ago

A sample of solid NH 4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (at

UNO [17]

Answer:

The answer to the question is

The equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains 0.26 atm.

Explanation:

To solve the question, we write out the chemical equation as follows

NH₄HS (s) ⇄ NH₃ (g) + H₂S (g)

From the above equation, it is observed that only the gaseous products contribute to the partial pressure

Kp =PNH₃·PH₂S where at Kp = 0.070 and PNH₃, PH₂S are the partial pressures of the gases

However since the number of moles of both gases are equal, therefore by Avogadro's law PNH₃ = PH₂S

Then PNH₃ = √(0.07) = PH₂S = 0.2645 atm. ≅ 0.26 atm.

5 0

3 years ago

How could you measure the average kinetic energy of a group of particles

dexar [7]

Answer:

Temperature measures the average kinetic energy of the particles in a substance. Thermal energy measures the total kinetic energy of the particles in a substance. The greater the motion of particles, the higher a substance's temperature and thermal energy.

Explanation:

8 0

3 years ago

Read 2 more answers

How many electrons can the fourth energy level can accommodate? a.2 b.8 c.16 d.32

OverLord2011 [107]

Answer:

8

Explanation:

8 0

3 years ago

A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of

Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0

3 years ago