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Likurg_2 [28]
4 years ago
15

Consider the reaction between HCl and O2: 4HCl grams + O2 - 2 H2O liquid + 2 Cl2 grams When 63.1 grams of HCl react with 17.2 g

of O2, 49.3 grams of Cl2 are collected. What is the limiting reactant?
Chemistry
1 answer:
raketka [301]4 years ago
6 0
The balanced equation for the above reaction is as follows;
4HCl + O₂ ---> 2H₂O + 2Cl₂
stoichiometry of HCl to O₂ is 4:1
number of HCl moles present - 63.1 g / 36.5 g/mol = 1.73 mol
number of O₂ moles present - 17.2 g / 32 g/mol = 0.54 mol
if HCl is the limiting reactant; then number of O₂ moles = 1.73/4 = 0.43 mol
There are 0.54 mol present and only 0.43 mol are required to react. Therefore HCl is the limiting reactant and O₂ is in excess.
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Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com
lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30  = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0
3 years ago
What happens to the do Broglle wavelength of an electron If its momentum ls doubled?​
algol13
Answer:
The De Broglie wavelength decreases when the momentum increases
Explanation:
The De Broglie wavelength of a particle (or any object) is given by

where
h is the Planck constant
p is the momentum of the object
As we can see, the wavelength is inversely proportional to the momentum of the object: therefore we can say that, if the momentum increases, the De Broglie wavelength will decrease.
8 0
3 years ago
An
Basile [38]

Answer:

The answer of this question is molecule

4 0
3 years ago
When a theory is disproved by further evidence, which of the following occurs?
morpeh [17]
Number 2 is the correct answer
3 0
3 years ago
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Swim ability is it materials ability to burn in the presence of
eduard

Answer: Flammability is a material's ability to burn in the presence of oxygen.

Explanation: Chemical properties can be observed only when the substance changes into one or more different substances through chemical reactions or transformations. One of the chemical properties is flammability.

Flammability is a material's ability to burn in the presence of oxygen.

Remember, oxygen doesn't burn. Precisely flammable substances obtain substances that burn. Oxygen remains an oxidizing agent, which means it supports the combustion process. Oxygen causes other objects to catch fire at low temperatures and burns hotter and faster. But oxygen itself does not burn. Consequently, if you at present deliver fuel and fire, adding oxygen will provide the fire.

Carbon dioxide is the result of combustion. An example can be seen in firewood in a fireplace. One of the chemical properties of carbon-based wood is having the ability to burn. Chemically the wood turns into carbon dioxide when it burns and leaves a residue of ash. Furthermore, this ash residue cannot be turned back into the wood. Chemical changes result in new substances.

Consider an example of a combustion reaction to methane gas:

Our balanced equation for methane combustion implies that every one CH₄ molecule reacts with two O₂ molecules. The product of combustion is one carbon dioxide molecule and two steam or water vapor molecules.

6 0
3 years ago
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