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Novosadov [1.4K]
3 years ago
5

Aluminum oxide forms when aluminum reacts with oxygen. 4Al(s) + 3O2(g) → 2Al2O3(s) A mixture of 82.49 g of aluminum (M.W.= 26.98

g/mol) and 117.65 g of oxygen (M.W. = 32.00 g/mol) is allowed to react. What mass of aluminum oxide (M.W. = 101.96 g/mol) can be formed?
Chemistry
2 answers:
Lena [83]3 years ago
7 0

Answer:

155.9 grams of aluminium oxide can be formed

Explanation:

Step 1: Data given

Mass of aluminium = 82.49 grams

Molar mass of aluminium = 26.98 g/mol

Mass of oxygen = 117.65 grams

Molar mass of oxygen = 32.0 g/mol

Molar mass of aluminium oxide = 101.96 g/mol

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate moles

Moles = mass / molar mass

Moles aluminium = 82.49 grams / 26.98 g/mol

Moles aluminium = 3.057 moles

Moles oxygen = 117.65 grams / 32.0 g/mol

Moles oxygen = 3.677 moles

Step 4: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Aluminium is the limiting reactant. It will completely be consumed (3.057 moles). Oxygen is in excess. There will react 3/4 * 3.057 = 2.293 moles

There will remain 3.677 - 2.293 = 1.384 moles

Step 5: Calculate moles aluminium oxide

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

For 3.057 moles Al we'll have 3.057/2 = 1.529 moles

Step 6: Calculate mass aluminium oxide

Mass aluminium oxide = 1.529 moles * 101.96 g/mol

Mass aluminium oxide = 155.9 grams

155.9 grams of aluminium oxide can be formed

NikAS [45]3 years ago
6 0

Answer:

155.4 g of Al₂O₃

Explanation:

The reaction is:

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

To determine the mass of aluminum oxide that is formed we need to know the limiting reagent. Let's calculate the moles of each by the molar mass

Mass / molar mass = Moles

82.49 g / 26.98 g/mol = 3.05 moles of Al

117.65 g / 32 g/mol = 3.67 moles of oxygen

Let's try the oxygen. Ratio is 3:4.

3 moles of O₂ need 4 moles of Al to react

Therefore 3.67 moles of O₂ will react with (3.67 . 4 )/3 = 4.90 moles

We only have 3.05 moles of Al, so the Al is the limiting reactant

Now, we work with stoichiometry

4 moles of Al can produce 2 moles of Al₂O₃

3.05 moles of Al will produce (3.05 .2) / 4 =  1.52 moles of Al₂O₃

We convert the moles to mass: 1.52 mol . 101.96 g / 1mol = 155.4 g

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fgiga [73]

Answer:

0.800 mol

Explanation:

We have the amounts of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with moles of the compounds involved.  

Step 1. <em>Gather all the information</em> in one place.

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

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===============

Step 2. Identify the <em>limiting reactant </em>

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<em>From C₃H₈:</em>

The molar ratio of CO₂: C₃H₈ is 3:1

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Moles of CO₂ = 12.0 mol CO₂

<em>From O₂</em>:

The molar ratio of CO₂: O₂ is 3:5.

Moles of CO₂ = 4.00 × ⅗

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O₂ is the limiting reactant because it gives the smaller amount of CO₂.

==============

Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.

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7 0
3 years ago
10. A small gold nugget has volume of 0.87 cm3. What is its mass if the density of gold is 19.3 g/cm3?
bulgar [2K]

Answer:

16.791 grams

Explanation:

The density formula is:

d=\frac{m}{v}

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

d*v=\frac{m}{v}*v

d*v=m

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

d= 19.3 g/cm^3\\v-0.87 cm^3

Substitute the values into the formula.

m=d*v

m= 19.3 g/cm^3*0.87 cm^3

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

m=19.3 g*0.87

m=16.791 g

The mass of the gold nugget is 16.791 grams.

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