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Norma-Jean [14]
3 years ago
6

Using the balanced chemical equation below. 2Al2O3 --> 4Al + 3O2 How many moles of oxygen are produced if 11.0 mol of Al are

produced?
Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
3 0

Answer:

8.25 moles of oxygen will be produced.

Explanation:

Given a balanced chemical equation;

2Al₂O₃ ------> 4Al + 3O₂

from the balanced chemical equation above, 3 moles of Oxygen are produced when 4 moles of Aluminum are produced.

4 moles of Al --------> 3 moles of Oxygen

11 moles of Al --------> ?

= \frac{(3 \ moles \ of \ O_2) \ \times \ (11 \ moles \ of \ Al)}{4 \ moles \ of \ Al} = 8.25 \ moles \ of \ O_2 \\\\

Therefore, 8.25 moles of oxygen will be produced.

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How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?
Tom [10]
The answer is:  " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂  = 2.0 mol CaCl₂ / L  ; 

Since: "M" = "Molarity" (measurement of concentration); 

                  = moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion:  1000 mL = 1 L . 

Given: 250 mL ;   

250 mL = ?  L  ?  ;  


250 mL * (1 L / 1000 L) =  (250/1000) L = 0.25 L . 
___________________________________________________________
 
(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol  = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ;  Convert to "g" (grams):

→ 0.50 mol CaCl₂  .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl  =  <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;  

→ Note the subscript, "2", in the " Cl₂ " ; 
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

40.08 g  +  2(35.45 g) = 

40.08 g  +  70.90 g = 110.98 g ;  round to 4 significant figures; 

                                 → round to 111 g/mol .
__________________________________________________________
So:

→  0.50 mol CaCl₂  = ? g CaCl₂  ? ; 

→  0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

                                             = (0.50) * (111 g) CaCl₂ ;

                                             =  55.5 g CaCl₂  ;

                                                → round to 2 significant figures; 

                                                →  56 g CaCl₂ .
___________________________________________________________
The answer is:  " 56 g CaCl₂ " .
___________________________________________________________
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1.2 x 10^10 mm^3 is equivalent to 12.0 m^3

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