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Pani-rosa [81]
3 years ago
10

What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?

Chemistry
1 answer:
jasenka [17]3 years ago
3 0

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

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I don’t know how to do this properly like I got an answer but I don’t know if it’s right
egoroff_w [7]

To calculate this, we will use the chemical equations as math equations and add them.

Firtly, we want the equation for the formation of CH₃CHO(g), so this will be the only product.

The reactants must be only the elements in their standard form, so C(g), O₂(g) and H₂(g). I would be more correct to use C(s), but since we odn't have information for this, we will assume it wants with C(g).

So, the reaction we want is:

C(g)+O_2(g)+H_2(g)\to CH_3CHO(g)

To balance the reaction, we can just do for eqach element separately, maintaining the coefficient of 1 on CH₃CHO(g):

\begin{gathered} 2C(g)+\frac{1}{2}O_2(g)+2H_2(g)\to CH_3CHO\mleft(g\mright) \\ \Delta H=? \end{gathered}

Now, we want to get to this equation adding the equations we want. We will apply the same operations to the enthalpies to get the enthalpy of formation.

The first given equation has the CH₃CHO(g), but it is on the left side and with coefficient of 2, so we need to invert the reaction and divided every coefficient by 2. The same operations have to be applied to the enthalpy, so the sign of the enthalpy will invert and it will be divided by 2:

\begin{gathered} 2CO_2(g)+2H_2O(l)\to CH_3CHO(g)+\frac{5}{2}O_2(g)_{} \\ \Delta H=\frac{2308.4kJ}{2}=1154.2kJ \end{gathered}

The second given equation has both C(g) and O₂(g), but since the third equation also has O₂(g), we will look just for C(g). We need 2 C(g), so we will need to doulbe the equation and its enthalpy:

\begin{gathered} 2C(g)+2O_2(g)\to2CO_2(g) \\ \Delta H=2\cdot-414.0kJ=-828.0kJ \end{gathered}

For the last, we will look into H₂(g) and since all the equations are balanced, O₂(g) will also be balanced by the end of it.

We need 2 H₂(g), so we don't need to do anything with this reaction:

\begin{gathered} 2H_2(g)+O_2(g)\to H_2O(l) \\ \Delta H=-597.4kJ \end{gathered}

Now, we add the equations:

\begin{gathered} \cancel{2CO_2\mleft(g\mright)}+\cancel{2H_2O\mleft(l\mright)}\to CH_3CHO(g)+\cancel{\frac{5}{2}O_2(g)}_{} \\ 2C(g)+\cancel{2O_2(g)}\to\cancel{2CO_2(g)} \\ 2H_2(g)+\cancel{O_2(g)}\to\cancel{H_2O(l)} \\ ------------------------------- \\ 2C(g)+\frac{1}{2}O_2(g)+2H_2(g)\to CH_3CHO(g) \end{gathered}

And we do the same with the enthalpies:

\begin{gathered} \Delta H=1154.2kJ+(-828.0kJ)+(-597.4kJ) \\ \Delta H=1154.2kJ-828.0kJ-597.4kJ \\ \Delta H=-271.2kJ \end{gathered}

This is the enthalpy for this reaction. To get the molar enthalpy of formation, we need to divide this value by the coefficient of CH₃CHO(g). Since this coefficient is 1, we have:

\Delta H_m=-\frac{271.2kJ}{1mol}=-271.2kJ\/mol

So, the molar enthalpy of formation given the data is -271.2 kJ/mol.

4 0
1 year ago
Describe the possible components of a buffer solution. (Select all that apply.)
AlexFokin [52]

Both D and G options. Weak base and its conjugate acid or weak acid and its conjugate base are the possible components of a buffer solution.

Explanation:

Buffer solution is the solution which gets easily dissolved in water and so called as "Aqueous solution".

Buffer solution is essentially made up of two components Known as:

i.) Weak base and its conjugate acid

ii.) Weak acid and its conjugate base

This weak acid and base solution is used to maintain the pH value of the solution in a balanced way.

When the weak acid or base solution is added to strong acid or base solution that is the way pH gets balanced .

In one word buffer solution is the solution which resists for the pH change when strong acids or bases are added.

7 0
3 years ago
There are moles of carbon present in 100 g of a
rodikova [14]

Answer: 3.33

Explanation:

4 0
2 years ago
What is the molar mass of calcium carbonate (CaCO3) ?
agasfer [191]

Answer: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

Explanation:

6 0
3 years ago
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Which of the substances are elements please help
ivanzaharov [21]

Answer:

Substances 1 and 2

Explanation:

an element only has 1 kind of atoms :3

6 0
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