Mg + __C12 = ____MgCl2
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Answer:
Explanation:
In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case ) and the anion is the atom at the right of the name (in this case
). With this in mind, the <u>formula would be</u>
.
Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> . So, we have as reagents:
The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:
With this in mind, the reaction would be:
I hope it helps!