Steochiochimistry! I don't get it!!?

Question

Can some one please help? Show balanced equaitions and cancellations.....

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).

Thanks for the help.....

Answer 1

Al = +3
O = -2

4Al + 3O2 = 2 Al2O3

then set up the ratio fraction

4 mole of Al makes 2 mole Al2O3

if you need 3.4 moles

3.4/X = 2/4
cross multipy and divide

(4*3.4)/2 = around 7 moles Al

which sound right because the 4 moles AL makes 2 mole of oxide. so to make 3.4 mole it would take a little more that 6. do the math i dont have a calculator and i hate the one on the computer.


n = 14
h = 1

this is how i do it.....

3.5/3 * 14 = about 14
3.5 * 1 = 3.5
about 14 + 3.5 = about 17.5

which sounds right because the reaction is 3 to 1.

hope it helps, im a chem student and i do short cuts and i they are hard to explain

Answer 2

The KEY to stoichiometry is that the chemical reaction relate the MOLAR amount of reactants (the stuff on the left side of the equation) to products (the stuff on the right side of the equation). You MUST do your comparisons of reaction components at the MOLE level, and, when necessary, go back to the mass level.

1. 4Al+3O2 -> 2Al2O3 the equation.
For every 2 moles of the oxide formed, 4 g-atoms (same as moles, except used for elements) of Al metal are needed. So we proceed by a proportion.
3.4 actual/2 equation Al2O3 = x actual/4 eqtn Al
solve for x = 6.8 g-atoms.

2. N2+3H2 -> 2 NH3 the equation.
First we convert 3.5 g to moles. Note that elemental hydrogen is a DIMER (H-H).
Moles = 3.5 g/2 g per mole = 1.75 moles
From the equation, for every 3 moles of H2, 2 moles of NH3 are produced. By proportion
1.75 mol/3 mol H2 = x mol/2 mole NH3.
Solve for x: 1.17 mole NH3 formed.
Go to grams: Mole wt NH3=17. Grams=17*1.17= 19.9 g (appx)

Answer 3

I came up with 19.3018 grams for the second question.


3.5 grams * 1 mole/2.02 grams
1.7 moles

Multiply by molar ratio

1.7/1 * 2/3

1.133333333 moles

Convert to grams

1.133333333 moles/1 * 17.031/1 moles

Result: 19. 3018 grams

Answer 4

Stoichiometry.

Answer 5

Can't even pronounce it!!