For the following reaction, if you have 13.2 g of CO and 42.7g of Fe2O3, which is the limiting reagent with regards to Fe production?

Fe2O3 (s) + 3 CO (g) _______> 2 Fe (s) + 3 CO2 (g)

Answers

Answer 1

Answer: [tex]CO[/tex] is the limiting reagent and [tex]Fe_2O_3[/tex] is the excess reagent.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Fe_2O_3=\frac{42.7g}{159.69g/mol}=0.267moles[/tex]

[tex]\text{Moles of} CO=\frac{13.2g}{28g/mol}=0.471moles[/tex]

The given balanced equation is :

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]

According to stoichiometry :

3 moles of [tex]CO[/tex] require = 1 mole of [tex]Fe_2O_3[/tex]

Thus 0.471 moles of [tex]CO[/tex] will require=[tex]\frac{1}{3}\times 0.471=0.157moles[/tex]  of [tex]Fe_2O_3[/tex]

As given amount of [tex]Fe_2O_3[/tex] is more than the required amount , it is the excess reagent.Thus [tex]CO[/tex] is the limiting reagent as it limits the formation of product.


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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
C6H12O6(aq) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

Calculate the volume of dry CO2 produced at body temperature (37°C) and 0.960 atm when 24.5 g of glucose is consumed in this reaction.

Answers

Answer:

[tex]\large \boxed{\text{21.6 L}}[/tex]

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

[tex]\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}[/tex]

(b) Moles of CO₂

[tex]\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}[/tex]

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

[tex]\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}[/tex]

The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.5 kJ/mol; and H2 O(g): Hf = –241.82 kJ/mol. How much heat is released by the combustion of 2 mol of methane? Use .

Answers

Answer:

The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol

Explanation:

The chemical reaction of the combustion of methane is given as follows;

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor

Where:

CH₄ (g): Hf = -74.6 kJ/mol

CO₂ (g): Hf = -393.5 kJ/mol

H₂O (g): Hf = -241.82 kJ/mol

Therefore, the combustion of 1 mole of methane releases;

-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol

Hence the combustion of 2 moles of methae will rellease;

2 × -802.54 kJ/mol or 1,605.08 kJ/mol.

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Just took the test

The element hydrogen has the highest specific heat of all elements. At room temperature, hydrogen's specific heat is 14.30 J/g •°C. If the temperature of a 340.0 g sample of hydrogen is to be raised by 30°C, how much energy will have to be added to the hydrogen?

Answers

Answer:

THE ENERGY NEEDED TO BE ADDED TO RAISE THE TEMPERATURE BY 30°C IS 145, 860 J

Explanation:

The energy needed can be calculated by using the heat formula, which is;

Heat = m c ΔT

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c = specific heat capacity of hydrogen at room temperature = 14.30 J/g°C

ΔT = change in temperature = 30°C

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Heat = 145860 J

The energy needed to be added to the hydrogen to bring the mass of 340 g to an increased temperature of 30°C is 145, 860 J or 145.86kJ

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Pipette and Burettes

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The two instruments that can be used for  

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Which is the dependent variable in this experiment?
Which is a controlled variable in this experiment?

Answers

In the experiment to measure the heat absorbed by a material placed in

the Sun, we have;

The independent variable is either the angle of insulation or the type of materialThe dependent variable is the temperature of the materialThe controlled variable is the type of material when the angle of insulation is the independent variable and vice versa

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The possible experiment is with regards to the factors that influence energy from the Sun

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The type of material and the angle of lighting (insulation) affects the absorption of heat

The independent variable is the variable that is the cause of the effect

measured in the experiment.

It is the variable that the researcher has control over or controls and

varies during the experiment to produce the measured effect.

In the above experiment, the independent variable are either ;

The angle of insulation or the type of material respectively

The dependent variable is the variable that is under test or that is being measured.

In the above experiment the dependent variable is the temperature of the material which can be measured

The controlled variable is the variable that remains the same or is held

constant during the experiment and can be used as reference to

compare the dependent variable.

In the experiment, the controlled variable are either;

The type of material when the angle of insulation is the independent variable or vice versa

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angle of light

temperature of material

type of material

Explanation:

E d g e 22 !

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P represents the atmosphere and Q represents the ocean.

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P represents the atmosphere and Q represents the ocean.

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