# 5 Solutions on Mole-to-Mole Stoichiometry Problems

Stoichiometry problems involving mole to mole calculations by using balanced chemical equations to determine the amount of one substance (in moles) required to react with a certain amount of another substance (also in moles).

In addition, every stoichiometry problem can be solved from the angle of mole-to-mole stoichiometry problem tips.

In mole-to-mole stoichiometry, there is need to be able to generate or deduce the mole-to-mole ratio, convert moles to reactants, volume or number of particles.

## Quick guidelines on solving mole to mole stoichiometry Problems

Mole-to-mole stoichiometry problems involve calculating the number of moles of a reactant or product involved in a chemical reaction.

There are several guidelines on solving mole to mole stoichiometry problems

*First, you should be able to write a balanced chemical equation for the reaction. This step is the first and most crucial to help to know the correct stoichiometry of the reaction.*

*Then Identify the known and unknown quantities in the balanced chemical equation. Most mole-to-mole stoichiometry problems involve the number of moles of one reactant or product, and you are asked to find the number of moles of another reactant or product.*

*Afterwards, convert the given quantity to moles, if necessary. If the given quantity is in grams or any other unit, you need to convert it to moles using the molecular weight or molar mass of the substance.*

*Next, apply the balanced chemical equation to set up a mole ratio between the known and unknown quantities. The stoichiometric coefficients in the balanced equation represent the mole ratios between the reactants and products.*

*Then the next step is to use the mole ratio gotten to calculate the unknown quantity. Multiply the number of moles of the known substance by the mole ratio to find the number of moles of the unknown substance.*

*Finally, convert the calculated quantity to the desired unit, if necessary. If the desired quantity is not in moles, you need to convert it to the desired unit using the molecular weight or molar mass of the substance.*

**Example 1**

Consider the equation Mg + Cl2 == MgCl2

If 3.8 moles of magnesium reacted, what mass of magnesium chloride would be produced?

Solution

The balanced chemical equation for the reaction between magnesium (Mg) and chlorine gas (Cl2) to form magnesium chloride (MgCl2) is:

2 Mg + Cl2 → 2 MgCl2

This equation shows that 2 moles of Mg react with 1 mole of Cl2 to produce 2 moles of MgCl2.

To calculate the mass of MgCl2 produced when 3.8 moles of Mg react, we need to use the mole ratio from the balanced equation.

From the equation, we can see that 2 moles of Mg react to produce 2 moles of MgCl2. Therefore, the mole ratio of Mg to MgCl2 is 2:2, or simply 1:1.

So, if 3.8 moles of Mg reacted, then we can expect 3.8 moles of MgCl2 to be produced.

To convert moles of MgCl2 to its mass, we need to use the molecular weight of MgCl2, which is approximately 95.2 g/mol.

Mass of MgCl2 produced = moles of MgCl2 produced × molecular weight of MgCl2 Mass of MgCl2 produced = 3.8 mol × 95.2 g/mol Mass of MgCl2 produced = 362.56 g

Therefore, when 3.8 moles of magnesium react with chlorine gas, the mass of magnesium chloride produced would be approximately 362.56 grams

**Example 2 **

Mg + ZnCl2 ==MgCl2 + Zn if 0.05 moles of Magnesium reacted with zinc chloride. Calculate mass of zinc produced

The balanced chemical equation for the reaction between magnesium and zinc chloride is:

Mg + ZnCl2 → MgCl2 + Zn

From the equation, we can see that the stoichiometry of the reaction is 1:1, meaning that 1 mole of magnesium reacts with 1 mole of zinc chloride to produce 1 mole of zinc.

We are given that 0.05 moles of magnesium reacted with zinc chloride. Therefore, we can conclude that 0.05 moles of zinc was produced.

To calculate the mass of zinc produced, we need to use the molar mass of zinc, which is 65.38 g/mol.

Mass of zinc produced = moles of zinc produced x molar mass of zinc Mass of zinc produced = 0.05 moles x 65.38 g/mol Mass of zinc produced = 3.27 g

Therefore, the mass of zinc produced is 3.27 g

**Example 3**

what volume of ammonia is evolved by reacting 5.6 moles of nitrogen with hydrogen according to this equation N2 + 3H2====2NH3

The balanced chemical equation for the reaction between nitrogen (N2) and hydrogen (H2) to produce ammonia (NH3) is

N2 + 3H2 → 2NH3

From the equation, we can see that 3 moles of hydrogen are required to react with 1 mole of nitrogen to produce 2 moles of ammonia.

1 mole of N2 ====2 moles of NH3

5.6 moles of N2=== X

X = 5.6 x 2 /1 =11.2 moles of Ammonia

Converting to volume

1 mole of NH3====22.4 dm3

11.2 moles of NH3=== x

X = 11.2 x 22.4 = 250.88 dm3 or 250.88 litres

**Example 4 **

Na+ 2H20 ===NaOH + 2H2 IF 2.8 moles of sodium reacted, what volume of hydrogen evolved

To calculate the volume of hydrogen evolved, we need to use the balanced chemical equation to determine the stoichiometric ratio between sodium and hydrogen. The balanced chemical equation is:

2 Na + 2 H2O → 2 NaOH + H2

From the equation, we can see that for every 2 moles of sodium (Na) that react, 1 mole of hydrogen gas (H2) is evolved. Therefore, to determine the volume of hydrogen evolved, we first need to calculate the number of moles of hydrogen produced.

Given that 2.8 moles of sodium (Na) reacted, we can use the stoichiometric ratio to determine the number of moles of hydrogen (H2) produced:

2 moles Na : 1 mole H2

2.8 moles Na : x moles H2

x = 1.4 moles H2

Now we can use the ideal gas law to calculate the volume of hydrogen gas produced at standard temperature and pressure (STP), which is defined as 0°C (273.15 K) and 1 atmosphere (1 atm) of pressure.

At STP, one mole of any gas occupies 22.4 dm3of volume. Therefore, the volume of hydrogen gas produced can be calculated as:

1.4 moles H2 x 22.4 dm3/mol = 31.36 dm3

Therefore, the volume of hydrogen gas evolved is 31.36 dm3 at STP

Hint = 1 litre = 1 dm3

**Example 5 **

CH4 + O2 ===CO2 + 2H2O IF 2.2 moles of methane burns in oxygen , calculate mass of carbon iv oxide

solution

To calculate the mass of carbon dioxide produced, we need to use the balanced chemical equation to determine the stoichiometric ratio between methane and carbon dioxide. The balanced chemical equation is:

CH4 + 2 O2 → CO2 + 2 H2O

From the equation, we can see that for every 1 mole of methane (CH4) that reacts, 1 mole of carbon dioxide (CO2) is produced. Therefore, to determine the number of moles of carbon dioxide produced, we need to calculate the number of moles of methane consumed.

Given that 2.2 moles of methane (CH4) reacted, we can use the stoichiometric ratio to determine the number of moles of carbon dioxide (CO2) produced:

1 mole CH4 : 1 mole CO2

2.2 moles CH4 : x moles CO2

x = 2.2 moles CO2

Now we can use the molar mass of carbon dioxide to calculate the mass produced. The molar mass of carbon dioxide is 44.01 g/mol.

mass of CO2 = moles of CO2 x molar mass of CO2 mass of CO2 = 2.2 moles x 44.01 g/mol mass of CO2 = 96.82 g

Therefore, the mass of carbon dioxide produced is 96.82 g.

**Example 6**

Problem: In the reaction 2H2 + O2 -> 2H2O, how many moles of water can be produced from 5 moles of hydrogen gas?

Solution:

- Write the balanced chemical equation: 2H2 + O2 -> 2H2O

- Determine the mole ratio of hydrogen gas to water from the balanced equation: 2 moles of H2 react to produce 2 moles of H2O

- Set up a proportion to solve for the unknown quantity (moles of H2O produced):

2 moles of H2O / 2 moles of H2 = x moles of H2O / 5 moles of H2

- Cross-multiply and solve for x:

2 moles of H2O * 5 moles of H2 = 2 moles of H2 * x moles of H2O

10 moles of H2O = 2x

x = 5 moles of H2O

Answer: 5 moles of water can be produced from 5 moles of hydrogen gas

in conclusion, solving mole to mole stoichiometry problems involve fewer steps but must be followed diligently.