Question

Write balanced chemical equations for the following reactions.

Answer 1

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.

Answer 2

❃ 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 ]

$\large{ \tt{❁ \: LET'S \: GET \: STARTED}} :$

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 : $\boxed{ \tt{CS₂ + 3O₂ ⟶ CO₂ + 2SO₂}}$

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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 : $\boxed{ \tt{Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O }}$

• 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! :)