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Sloan [31]
3 years ago
10

Label each Formula In the Chemical equation Below As either A reactant or a product

Chemistry
1 answer:
lara [203]3 years ago
4 0
In chemical reactions, the reactants are consumed to create the products, and (assuming this is a normal chemical reaction) the reactants are located to the left of the arrow while the products are located to the right of the arrow. So in this case, Fe would be a reactant, S would be a reactant, and the FeS would be a product.

Hope this helps
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What is the Net Ionic equation for this chemical reaction: FeBr2+Na2S=FeS+2NaBr​
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Answer: Fe<em>(aq)</em>+S<em>(aq)</em>=FeS<em>(s)</em>

Explanation: The Sodium and Bromine are spectator ions because they don't react with anything, you can see this by writing the ionic equation like so:

1.) Molecular formula (given): FeBr2 (aq)+Na2S (aq)= FeS(s)+2NaBr(aq)

Each dissolved FeBr2 breaks up into one Fe with a charge of 2+ and two Br with a negative charge. This gives you:

Fe(aq)+ 2Br(aq)+Na2S(aq)=FeS(s)+2NaBr

2.) Now repeat what was shown with the other compounds in the given molecular formula, and pay attention to the states that each ion is in (solid, liquid, aqueous, gas) because this will give you the ionic equation, which from there you can get rid of any ions that don't change amount or state.

3.) Ionic formula: Fe(aq)+ <u>2Br(aq)</u>+<u>2 Na(aq)</u>+S (aq)=FeS(s)+<u>2 Na(aq)+2Br(aq)</u>

4.)When you've derived a total ionic equation (above), you'll  find that some ions appear on both sides of the equation in equal numbers. For example, in this case two Na cations and two Br anions appear on both sides of the total ionic equation. What does this mean? It means these ions don't participate in the chemical reaction. They're present before and after the reaction. Nothing happens to them. So those are removed and you're left with the net ionic: Fe(aq)+S(aq)=FeS(s)

Hope this helps :)

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NO \longrightarrow N_2O_2 \longrightarrow N_2O \longrightarrow N2

Explanation:

The intermediates are the products of all the steps of the reaction pathway, with the exception of the last one. So the intermediates will be:

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