How do you balance $A l + F_{2} \to A l F_{3}$ $Al + F_2 -> AlF_3$ ?

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How do you balance $A l + F_{2} \to A l F_{3}$ $Al + F_2 -> AlF_3$ ?

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Answer 1 · 2016-02-28T07:37:03

$A l + F_{2} \to A l F_{3}$ $Al + F_2 rarrAlF_3$
here F is unbalanced : on right hand side(product side) there are 3 F and on the left hand side there is 2F

use the L.C.M of the two numbers to balance on both sides
here L.C.M of (2,3)=6
so we try to get six F on both sides

$A l + 3 F_{2} \to 2 A l F_{3}$ $Al + 3F_2 rarr 2AlF_3$
now we need to balance Al
$2 A l + 3 F_{2} \to 2 A l F_{3}$ $2Al +3F_2 rarr2AlF_3$

Answer 2 · 2016-02-28T07:39:14

You balance it stoichiometrically: GARBAGE IN MUST EQUAL GARBAGE OUT.

Explanation:

A chemical reaction ALWAYS CONSERVES MASS AND CHARGE:

$2 A l (s) + 3 F_{2} (g) \to 2 A l F_{3} (s)$ $2Al(s) + 3F_2(g) rarr 2AlF_3(s)$ .

Is mass conserved in this representation? How? If you have to do calculations with this formula, to make it a little easier on yourself you could write:

$A l (s) + \frac{3}{2} F_{2} (g) \to A l F_{3} (s)$ $Al(s) + 3/2F_2(g) rarr AlF_3(s)$

What does $\frac{1}{2} F_{2} (g)$ $1/2F_2(g)$ represent in $grams$ $"grams"$ ?

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How do you balance $A l + F_{2} \to A l F_{3}$ $Al + F_2 -> AlF_3$ ?

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