Respuesta :
❃ The following points should be kept in mind to write and balance a chemical equation :
Step 1 : Write the molecular formula of all the reactants and products correctly.
Step 2 : Separate reactants and products by a sign of arrow. If reactants or products are more than one, connect them by a sign of a plus.
Step 3 : Balance the atoms of O and H at last [ The atoms used at many places in an equation should be balanced at last ]. For balancing , the number should be added as coefficient i.e in the front of the molecules.
[ Remember those substance that take part in a chemical reaction are called reactants. Likewise , those substances which are formed after a chemical reaction are called products ]
[tex] \large{ \tt{❁ \: LET'S \: GET \: STARTED}} : [/tex]
1. Carbon disulfide + Oxygen gas gives carbon dioxide + Sulfur dioxide.
Step 1 : The molecular formula of carbon disulfide is CS₂ , molecular formula of Oxygen gas is 0₂ [ Since oxygen is a diatomic element ] molecular formula of carbon dioxide is CO₂ and molecular formula of sulfur dioxide is SO₂.
Step 2 : CS₂ + O₂ ⟶ CO₂ + SO₂
Step 3 : In the reactant side , there is two ' S ' but on the other side , there is one ' S '. So , add 2 as a coefficient before S on the product side. Now , There are two ' O ' in the reactant side but six ' O ' in the product side. So , add 3 as a coefficient before O on the reactant side. Now , there are equal atom of C , S and O on both sides
i.e CS₂ + 3O₂ ⟶ CO₂ + 2SO₂
Answer : [tex] \boxed{ \tt{CS₂ + 3O₂ ⟶ CO₂ + 2SO₂}}[/tex]
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2. Silver + nitric acid gives silver nitrate + nitrogen dioxide + water
Step 1 : The molecular formula of Silver is Ag, molecular formula of nitric acid is HNO₃ , molecular formula of Silver nitrate is Ag ( No₃ ) , molecular formula of nitrogen dioxide is NO₂ and molecular formula of water is H₂O.
Step 2 : Ag + HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O
Step 3 : In the reactant side , There is one ' H ' but on the other side , there are two ' H '. Now add 2 before H on the reactant side. There are equal atom of ' Ag ' , ' H ' , ' N ' , and ' O '.
i.e Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O
Answer : [tex] \boxed{ \tt{Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O }}[/tex]
- Yay! We're done ! :)
- The last step is a bit more confusing I guess. So , which balancing , count the atoms in following ways :
- The number written at the right lower corner of an atom is counted for that atom only. For example : In MgSO₄ , there are one ' Mg ' , one ' S ' and four ' O '
- The number written at the right lower corner of a bracket is for all the atoms enclosed in the bracket. For example : In Al₂ ( SiO₃ ) has two Al , three ' S ' and nine ' O '.
- The coefficient number is for all the atoms of the molecule. For example , in 2 Al ₂( SiO₃ )₃ , there are four ' Al ' , six ' Al ' and eighteen ' O '.
- Hope this helps! :)
Answer:
a) CS₂ + 3O₂ → CO₂ + 2SO₂
b) Ag + 2HNO₃ → AgNO₃ + NO₂ + H₂O
General Formulas and Concepts:
Atomic Structure
- Reading a Periodic Table
- Reactions rxn
- Compounds
Aqueous Solutions
- States of matter
Explanation:
a)
Carbon (C) has a charge of ±4 on the periodic table. Sulfur (S) has a charge of -2 on the periodic table. disulfide would be 2 times sulfur, so S₂. Oxygen (O) has a charge of -2 also on the periodic table:
Carbon disulfide = CS₂
Oxygen gas = O₂
Carbon dioxide (remember this by now) = CO₂
Sulfur dioxide = SO₂
Write out our unbalanced chemical rxn:
CS₂ + O₂ → CO₂ + SO₂
We see that our number of sulfurs is unbalanced. We need to balance out the sulfurs:
CS₂ + O₂ → CO₂ + 2SO₂
Now we see that our sulfurs are balanced and that we have 2 oxygens on the reactant side and 6 on the product side. Balance out the oxygens on the reactant side to equal 6:
CS₂ + 3O₂ → CO₂ + 2SO₂
And that is our balanced rxn.
b)
Recall that the polyatomic ion nitrate is (NO₃⁻). Silver (Ag) has a charge of +1 on the periodic table:
Silver = Ag
Nitric Acid (strong acid - remember): HNO₃
Silver Nitrate = AgNO₃
Nitrogen dioxide: NO₂
Water (universal): H₂O
Let's write our unbalanced rxn:
Ag + HNO₃ → AgNO₃ + NO₂ + H₂O
We see that there are 2 nitrogens on the products side and 1 on the reactant side. We need to balance this out by increasing the reactant side:
Ag + 2HNO₃ → AgNO₃ + NO₂ + H₂O
We see that now our nitrogens are balanced out. Moving onto hydrogens, we see that there are 2 hydrogens on each side, balanced out.
Moving onto oxygens, we can also see that there are 6 oxygens on both sides. Last but not least, we have 1 silver on both sides.
Therefore, that would be our balanced rxn.