Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum ar

Question

Problem #1:
Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum are needed to form 3.4 moles of Aluminum oxide?
Hint: This is a simple ratio problem. Just use the molar ratio from the balanced equation. This was covered in 5.09. Mass is not involved.

Problem #2:
Determine the number of grams of NH3 produced by the reaction of 3.5g of hydrogen gas with sufficient nitrogen gas. (Check for diatomic elements).
Can U Show me how 2 balence and how 2 and what 2 cancel.
PLEASE I NEED YOUR HELP!

Answer 1

Problem #1:
First, think of what the molecular formular for aluminum oxide would be. The aluminum ion has a 3+ charge, and the oxygen ion has a 2- charge. You want the entire compound to be neutral, so the aluminum oxide compound needs to have two aluminum ions and three oxygen ions ([2 x +3] + [3 x -2] = 0, a neutral charge.) So the unbalanced equation for the formation of aluminum oxide would be:
Al(s) + O2(g) = Al2O3(s)
In order to balance, you need the same number of elements on each side of the equation. We need 2 Al one the left side and 3 O on the left side. But wait! Oxygen is diatomic, meaning in its natural state, it occurs as a molecule consisting of two atoms. So heres what you do: put 2 in front of Al and put 3/2 in front of O2.
2Al(s) + 3/2O2(g) = Al2O3(s)
Now in order to get rid of the fraction, multiply the entire equation by the denominator (in this case, 2).
4Al + 3O2 = 2Al2O3
Check to see that it's balanced. We have four Al on each side, and six O on each side, so it is balanced.
Now, we want 3.4 moles of Al2O3, our product. In order to find out how many moles of Al we need, we must multiply by the ratios, making sure units cancel. In other words, we have 4 moles Al for every 2 moles of Al2O3 in our balanced equation, so:
3.4mol Al2O3 x 4mol Al/2mol Al2O3 = 6.8mol Al
(Notice we have moles Al2O3 on top and on bottom, so they cancel to give us Al.)

Problem #2:
Similar problem. Both hydrogen and nitrogen are diatomic, so our unbalanced equation is:
H2(g) + N2(g) = NH3(g)
We need the same number of elements on each side, so if we put 2 in front of NH3 and 3 in front of H2, it will be balanced.
3H2(g) + N2(g) = 2NH3(g)
Now, convert 3.5g H2 to moles. Remember, hydrogen is diatomic, so:
3.5g x mol/(2 x 1.008)g = 1.736mol H2.
Now we use the ratios from our balanced equation.
1.736mol H2 x 2mol NH3/3mol H2 = 1.157mol NH3.
Now we need to calculate the molecular mass of ammonia by using the values from the periodic table.
(1 x 14.01) + (3 x 1.008) = 17.034g/mol
Now use the number of moles calculated earlier and multiply by the molecular mass.
1.157mol x 17.034g/mol = 19.708g NH3

Answer 2

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RE:
Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum ar
Problem #1:
Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum are needed to form 3.4 moles of Aluminum oxide?
Hint: This is a simple ratio problem. Just use the molar ratio from the balanced equation. This was covered in 5.09. Mass is not...

Answer 3

Aluminum Oxide Reaction

Answer 4

#1
The balanced equation will be:

4 Al + 3 O2 --> 2 Al2O3

Since the ratio of aluminum to aluminum oxide is 2:1, just multiply the moles of Al2O3 by 2 to get the moles of Al needed.

#2
The reaction is between H2 and N2 to give NH3.
H2 + N2 --> NH3

To balance it, since you have N2 on the left, you'll form 2 NH3 on the right. That gives you 6 H on the right, so you'll need 3 H2 on the left. The balanced equation is:

3 H2 + N2 --> 2 NH3

So get to the answer, first convert the grams of H2 into moles by dividing the the molar mass of H2. Then, use the coefficients of the equation to convert moles of H2 into moles of NH3. Finally, use the molar mass of NH3 to convert moles of NH3 into grams of NH3.

Answer 5

Safely & Permanently Remove Moles, Warts and Skin Blemishes

Answer 6

Only if you promise to stop using IM-speak...

1.)
4Al + 3O2 --> 2Al2O3

So double the moles of aluminum oxide you're trying create, so you need 6.8 moles of Al to run the reaction to the point you want it.

2.)
N2 + 3H2 --> 2NH3

3.5 g H2/2 g/mol = 1.75 moles

1.75 moles H2 * 2 moles NH3/3 moles H2 = 7/6 moles NH3

7/6 moles NH3 * 17 g/mol = 19.8 grams

Answer 7

4Al + 3O2 => 2Al2O3

2 moles of Al2O3 formed from 4 moles of Al
3.4 moles of Al2O3 forms ???
(cross multiply)
=(3.4 X 4) divided by 2
=6.8 moles of Al

2) 2N + 3H2 => 2NH3

3 moles of H2 is needed to form 3 moles of NH3
1.75* moles is needed to form ????? (* = 3.5/2, because hydrogen is diatomic)

(cross multiply)
1.75 X 3 divided by 3
=1.75 moles
mass= moles X relative atomic mass
mass= 1.75 X 17 (N=14 , H=1)
mass= 29.75g