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stepan [7]
2 years ago
7

Using chemical equation to find moles of product from moles of reactant

Chemistry
1 answer:
Zolol [24]2 years ago
5 0

Answer:

0.21 mole of C8H18.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Next, we shall determine the number of mole of C8H18 that reacted and the number of mole water (H2O) produced from the balanced equation.

This is illustrated below:

From the balanced equation above,

2 moles of C8H18 reacted to produce 18 moles of H2O.

Finally, we shall determine the number of mole C8H18 needed to produce 1.90 moles of H2O.

This can be obtained as follow:

From the balanced equation above,

2 moles of C8H18 reacted to produce 18 moles of H2O.

Therefore, Xmol of C8H18 will react to produce 1.90 moles of H2O i.e

Xmol of C8H18 = (2 × 1.90)/18

Xmol of C8H18 = 0.21 mole

Thus, 0.21 mole of C8H18 is needed for the reaction.

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__________________________________________________________

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

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

                  = moles of solute per L {"Liter"} of solution.
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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 . 
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We have: 0.50 mol CaCl₂ ;  Convert to "g" (grams):

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

From the Periodic Table of Elements:

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</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 .
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The answer is:  " 56 g CaCl₂ " .
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