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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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Answer:

2.33g of iron (iii) chloride

50.0 mL of 5.00 M of sodium phosphate

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mol = conc × vol = 0.5 × 50/1000 = 0.025 mol Na3PO4

from the equation:

1 mol of Na3PO4 reacts with 1 mol FeCl3 = 3 mol of NaCl

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3 0
2 years ago
Describe any new ways you can change the polarity of the three-atom molecule.
yan [13]

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5 0
3 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
The total number of sodium atoms in 46.0 grams of solution is?
USPshnik [31]
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7 0
3 years ago
Read 2 more answers
Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa
Lapatulllka [165]

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason: 

If you look at the combining masses of X and Y in each of the two compounds, 

The first compound contains 0.25g of X combined with 0.75g of Y 
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The second compound contains 0.33 g of X combined with 0.67 g of Y 
so the ratio (by mass) of X to Y = 1 : 2 

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

The first compound will then contain 4g of X and 12g of Y 
The second compound will then contain 6g of X and 12g of Y 

<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6 
<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

8 0
3 years ago
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