## Chemical Equation - Stoichiometry and Basic Chemistry Calculations

## Video Lesson

Solve any problem involving stoichiometry using 5 simple steps.

The best way to learn this topic is by working through practice problems. We will look at solving one practice question in this post.

*Watch the video for detailed step-by-step explanation of this and another question.*

## Question

Calculate the **mass of NaCl** produced when **2g of sodium** reacts with excess **chlorine**.

## STEP 1 : Write a balanced chemical equation

The first step is to determine the **balanced chemical equation** of the reaction.

Identify the reactants and products from the question.

From Question 1:

Reactants – **sodium** and **chlorine**

Products – **sodium chloride**

Therefore, the chemical equation would be:

Na (s) + Cl_{2} (g) â†’ NaCl (s)

Now we need to balance the number of atoms of each element on the left and right sides of the chemical equations.

Left:

Na – 1

Cl – 2

Right:

Na – 1

Cl – 1

To balance the number of chlorine atoms, we need to add “**2**” in front of **NaCl**.

Na (s) + Cl_{2} (g) â†’ **2**NaCl (s)

Now we have:

Left:

Na – 1

Cl – 2

Right:

Na – 2

Cl – 2

To balance the number of sodium atoms, we need to add “**2**” in front of **Na**.

**2**Na (s) + Cl_{2} (g) â†’ 2NaCl (s)

Now we have:

Left:

Na – 2

Cl – 2

Right:

Na – 2

Cl – 2

Now the chemical equation is finally balanced. We have the same number of sodium and chlorine atoms on both sides of the chemical equation.

## Question

Calculate the **mass of NaCl** produced when **2g of sodium** reacts with excess chlorine.

## STEP 2 : Find mole ratio of species involved

2Na (s) + Cl_{2} (g) â†’ 2NaCl (s)

Now that we have a balanced chemical equation, we can analyse it quantitatively.Â

- The number in front of each species is known as the
**coefficient**of the species. - These coefficients tell us the
**ratio of moles**of the species involved in the chemical reaction. - From the equation we know that:

**2 mol** of sodium reacts with **1 mol** of chlorine to produce **2 mol** of sodium chloride

When deciding which ratio of moles to find, we need to carefully examine the question and look for two things:

- Quantity of species to be found (
**mass of NaCl**) - Quantity of species given (
**2g of sodium**)

Now we know that we need to find the ratio of moles of sodium chloride,Â **NaClÂ **to sodium,Â **Na**.

From the quantitative analysis we did earlier, we know that:

moles of NaCl : moles of Na

= 2 : 2

Both “**2**“s are theÂ **coefficientsÂ **of each species in the balanced chemical equation.

**2**Na (s) + Cl_{2} (g) â†’ **2**NaCl (s)

We can simplify this by dividing both numbers by 2 to get:

moles of NaCl : moles of Na

= 2 : 2

= 1 : 1

## Question

Calculate the **mass of NaCl** produced when **2g of sodium** reacts with excess chlorine.

## STEP 3 : Calculate actual number of moles of species with given quantity

- The question tells us thatÂ
**2g of sodiumÂ**has reacted.Â - This is the given quantity (
**mass of sodium**).Â - We can use this quantity to find the number ofÂ
**moles of sodiumÂ**reacted.

There are four equations for number of moles which must be committed to memory:

m. m. = molar mass

m. v. = molar volume

# of particles = number of particles

N_{A} = Avogardro’s constant = 6.02 x 10^{23}

M = molarity

V = volume

Since we are given the mass of sodium, we would choose the mole formula that includes mass in it:

Â

Â

## Question

Calculate the **mass of NaCl** produced when **2g of sodium** reacts with excess chlorine.

## STEP 4 : Calculate actual number of moles of species with unknown quantity.

Now that we have the **actual number of moles** of **sodium**, we can use the **mole ratio** we determined in **STEP 2** to calculate the **actual number of moles** of **sodium chloride** formed:

moles of NaCl : moles of Na

= 1 : 1

= x : 0.0870

Using the fraction form we have:

Â

## Question

Calculate the **mass of NaCl** produced when **2g of sodium** reacts with excess chlorine.

## STEP 5 : Calculate unknown quantity

From **STEP 4**, we have the **number of moles** of **sodium chloride** formed.

Since we need to find the **mass** of sodium chloride formed, we should choose the formula that relates mass to number of moles:

The **molar mass** of **sodium chloride** has the same value as its **relative formula mass**.

This can be found by adding the **relative atomic mass** of all the ions in the formula unit NaCl:

Â

Â

Using these **5 steps** and the **four mole formulas**, you can find any quantity **(mass/volume/number of particles/molarity)** of a species as long as the quantity of another species in the chemical equation is given.

## Practice Questions

- Calculate the volume of oxygen gas needed at RTP to react with 4g of lithium to form lithium oxide.
**Ans: 3.43 dm**^{3} - Calculate the number of atoms of magnesium that will produce 3.5g of zinc when magnesium reacts with zinc chloride to form zinc and magnesium chloride.
*Ans: 3.24 x 10*^{22}atoms - Calculate the volume of carbon dioxide produced at RTP when 8.4g of carbon is reacted with excess iron(III) oxide to form iron and carbon dioxide.
*Ans: 16.8Â**dm*^{3}

### Get in touch to join my class

60108280139