How do you balance $N a + M g F_{2} \to N a F + M g$ $Na + MgF_2 -> NaF + Mg$ ?

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How do you balance $N a + M g F_{2} \to N a F + M g$ $Na + MgF_2 -> NaF + Mg$ ?

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Answer 1 · 2015-12-07T11:40:41

$2 N a + M g F_{2} \to 2 N a F + M g$ $2Na+MgF_2 rarr 2NaF + Mg$

Explanation:

The $2$ $2$ F's on the left side ( $M g F_{2}$ $MgF_color(red)(2)$ ) imply the that right side must have $2$ $2$ F's (or a multiple of $2$ $2$ ).

The only F on the right side is in NaF
so to have $2$ $2$ F's on the right side we will need $2$ $2$ NaF's.
$XXX 2 N a F + M g$ $color(white)("XXX")color(red)(2NaF)+Mg$

This gives us $2$ $2$ Na's on the right side ( $2 N a F$ $color(blue)(2Na)F$ )
which implies we will need $2$ $2$ Na's on the left side.
$XXX 2 N a + M g F_{2} \to 2 N a F + M g$ $color(white)("XXX")color(blue)(2)Na + MgF_2rarr color(blue)(2Na)F + Mg$

At this point the transformation becomes balanced:
On both sides we have:

$2 N a$ $2 Na$ 's
$1 M g$ $1 Mg$ , and
$2 F$ $2 F$ 's

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How do you balance $N a + M g F_{2} \to N a F + M g$ $Na + MgF_2 -> NaF + Mg$ ?

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How do you balance Na+MgF2→NaF+MgNa + MgF_2 -> NaF + Mg?

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How do you balance $N a + M g F_{2} \to N a F + M g$ $Na + MgF_2 -> NaF + Mg$ ?