Question #988dd

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Question #988dd

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Answer 1 · 2017-04-02T17:33:09

$Z n C l_{2} + N a_{2} C O_{3} \to Z n C O_{3} + 2 N a C l$ $ZnCl_2 + Na_2CO_3 rarr ZnCO_3 + 2 NaCl$

Explanation:

$Z n C l_{2} + N a_{2} C O_{3} \to Z n C O_{3} + N a C l$ $ZnCl_2 + Na_2CO_3 rarr ZnCO_3 + NaCl$

This is what I do...
I write that all elements on the reactant side as well as those on the product side. Not forgetting to split them when there is an addition.

Reactants

$Z n = 1$ $Zn=1$
$C l = 2$ $Cl=2$
$+$ $+$
$N a = 2$ $Na=2$
$C = 1$ $C=1$
$O = 3$ $O=3$

Products

$Z n = 1$ $Zn=1$
$C = 1$ $C=1$
$O = 3$ $O=3$
$+$ $+$
$N a = 1 \times 2 = 2$ $Na=1xxcolor(red)2=2$
$C l = 1 \times 2 = 2$ $Cl=1xxcolor(red)2=2$

From the above, we see that $N a$ $Na$ is $2$ $2$ on the reactant side and $1$ $1$ on the product side so we multiply by $2$ $color(red)2$ to make it equal. When you do this, it automatically affects $C l$ $Cl$ because it forms a compound.

Check to see if all elements are balanced then rewrite the equation.

$Z n C l_{2} + N a_{2} C O_{3} \to Z n C O_{3} + 2 N a C l$ $ZnCl_2 + Na_2CO_3 rarr ZnCO_3 + color(red)2 NaCl$

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Question #988dd

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