If $378.2$ Grams Of $Cd$ Are Reacted, How Many Grams Of $AgNO_3$ Will Also Be Reacted In The Reaction:$\[ Cd + 2AgNO_3 \rightarrow Cd(NO_3)_2 + 2Ag \\]

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Introduction

Chemical reactions are the foundation of chemistry, and understanding the relationships between reactants and products is crucial in various fields, including chemistry, physics, and engineering. One of the fundamental concepts in chemistry is stoichiometry, which deals with the quantitative relationships between reactants and products in chemical reactions. In this article, we will explore the concept of stoichiometry and how to balance chemical equations using the example of the reaction between cadmium (Cd) and silver nitrate (AgNO3).

What is Stoichiometry?

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It involves the calculation of the amounts of reactants and products required or produced in a chemical reaction. Stoichiometry is essential in chemistry as it helps us understand the limitations of a reaction, predict the yield of a product, and design experiments to produce a specific amount of a substance.

Balancing Chemical Equations

A chemical equation is a representation of a chemical reaction, showing the reactants on the left side and the products on the right side. However, most chemical equations are not balanced, meaning that the number of atoms of each element is not the same on both sides of the equation. Balancing a chemical equation involves adding coefficients (numbers in front of the formulas of reactants or products) to ensure that the number of atoms of each element is the same on both sides of the equation.

The Reaction between Cadmium (Cd) and Silver Nitrate (AgNO3)

The reaction between cadmium (Cd) and silver nitrate (AgNO3) is a classic example of a single displacement reaction, where one element displaces another element from a compound. The balanced chemical equation for this reaction is:

Cd+2AgNO3→Cd(NO3)2+2Ag{ Cd + 2AgNO_3 \rightarrow Cd(NO_3)_2 + 2Ag }

In this reaction, cadmium (Cd) reacts with two moles of silver nitrate (AgNO3) to produce one mole of cadmium nitrate (Cd(NO3)2) and two moles of silver (Ag).

Calculating the Amount of AgNO3 Required

To calculate the amount of AgNO3 required to react with 378.2 grams of Cd, we need to use the concept of stoichiometry. The balanced chemical equation shows that 1 mole of Cd reacts with 2 moles of AgNO3. We can use the molar mass of Cd and AgNO3 to calculate the number of moles of Cd and AgNO3 required.

Molar Mass of Cd and AgNO3

The molar mass of Cd is 112.41 g/mol, and the molar mass of AgNO3 is 169.87 g/mol.

Calculating the Number of Moles of Cd

To calculate the number of moles of Cd, we can use the formula:

n=mM{ n = \frac{m}{M} }

where n is the number of moles, m is the mass of Cd (378.2 g), and M is the molar mass of Cd (112.41 g/mol).

nCd=378.2 g112.41 g/mol=3.36 mol{ n_{Cd} = \frac{378.2 \, g}{112.41 \, g/mol} = 3.36 \, mol }

Calculating the Number of Moles of AgNO3 Required

Since 1 mole of Cd reacts with 2 moles of AgNO3, we can calculate the number of moles of AgNO3 required by multiplying the number of moles of Cd by 2:

nAgNO3=2×nCd=2×3.36 mol=6.72 mol{ n_{AgNO_3} = 2 \times n_{Cd} = 2 \times 3.36 \, mol = 6.72 \, mol }

Calculating the Mass of AgNO3 Required

To calculate the mass of AgNO3 required, we can use the formula:

m=n×M{ m = n \times M }

where m is the mass of AgNO3, n is the number of moles of AgNO3 (6.72 mol), and M is the molar mass of AgNO3 (169.87 g/mol).

mAgNO3=6.72 mol×169.87 g/mol=1137.51 g{ m_{AgNO_3} = 6.72 \, mol \times 169.87 \, g/mol = 1137.51 \, g }

Therefore, 1137.51 grams of AgNO3 will be required to react with 378.2 grams of Cd.

Conclusion

In this article, we have explored the concept of stoichiometry and how to balance chemical equations using the example of the reaction between cadmium (Cd) and silver nitrate (AgNO3). We have calculated the amount of AgNO3 required to react with 378.2 grams of Cd using the concept of stoichiometry. The balanced chemical equation shows that 1 mole of Cd reacts with 2 moles of AgNO3, and we have used the molar mass of Cd and AgNO3 to calculate the number of moles of Cd and AgNO3 required. Finally, we have calculated the mass of AgNO3 required to react with 378.2 grams of Cd.

References

  • CRC Handbook of Chemistry and Physics, 97th ed., CRC Press, 2016.
  • Chemistry: An Atoms First Approach, 2nd ed., Steven S. Zumdahl, Cengage Learning, 2014.
  • General Chemistry: Principles and Modern Applications, 11th ed., John W. Hill, William S. O'Brien, 2015.

Q: What is the difference between a balanced chemical equation and an unbalanced chemical equation?

A: A balanced chemical equation is an equation in which the number of atoms of each element is the same on both sides of the equation. An unbalanced chemical equation is an equation in which the number of atoms of each element is not the same on both sides of the equation.

Q: How do I balance a chemical equation?

A: To balance a chemical equation, you need to add coefficients (numbers in front of the formulas of reactants or products) to ensure that the number of atoms of each element is the same on both sides of the equation.

Q: What is the concept of stoichiometry?

A: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It involves the calculation of the amounts of reactants and products required or produced in a chemical reaction.

Q: How do I calculate the number of moles of a substance?

A: To calculate the number of moles of a substance, you can use the formula:

n=mM{ n = \frac{m}{M} }

where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance.

Q: What is the difference between a mole and a gram?

A: A mole is a unit of measurement that represents 6.022 x 10^23 particles (atoms or molecules), while a gram is a unit of mass. The molar mass of a substance is the mass of one mole of that substance.

Q: How do I calculate the mass of a substance required to react with a given amount of another substance?

A: To calculate the mass of a substance required to react with a given amount of another substance, you need to use the concept of stoichiometry. You can calculate the number of moles of the substance required, and then multiply that number by the molar mass of the substance to get the mass required.

Q: What is the significance of balancing chemical equations?

A: Balancing chemical equations is important because it ensures that the number of atoms of each element is the same on both sides of the equation. This is crucial in predicting the yield of a product and designing experiments to produce a specific amount of a substance.

Q: Can you give an example of how to balance a chemical equation?

A: Yes, let's consider the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O):

H2+O2→H2O{ H_2 + O_2 \rightarrow H_2O }

To balance this equation, we need to add coefficients to ensure that the number of atoms of each element is the same on both sides of the equation. The balanced equation is:

2H2+O2→2H2O{ 2H_2 + O_2 \rightarrow 2H_2O }

Q: What is the difference between a limiting reactant and an excess reactant?

A: A limiting reactant is the reactant that is present in the smallest amount, and it determines the amount of product that can be formed. An excess reactant is the reactant that is present in excess of the amount required to react with the limiting reactant.

Q: How do I determine the limiting reactant in a reaction?

A: To determine the limiting reactant in a reaction, you need to calculate the number of moles of each reactant required to react with the other reactant. The reactant that is present in the smallest amount is the limiting reactant.

Q: What is the significance of determining the limiting reactant?

A: Determining the limiting reactant is important because it helps you predict the yield of a product and design experiments to produce a specific amount of a substance.

Q: Can you give an example of how to determine the limiting reactant in a reaction?

A: Yes, let's consider the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O):

2H2+O2→2H2O{ 2H_2 + O_2 \rightarrow 2H_2O }

To determine the limiting reactant, we need to calculate the number of moles of each reactant required to react with the other reactant. Let's say we have 2 moles of H2 and 1 mole of O2. The balanced equation shows that 2 moles of H2 react with 1 mole of O2 to form 2 moles of H2O. Therefore, H2 is the limiting reactant, and O2 is the excess reactant.

Q: What is the difference between a mole ratio and a mass ratio?

A: A mole ratio is the ratio of the number of moles of two substances, while a mass ratio is the ratio of the mass of two substances.

Q: How do I calculate a mole ratio?

A: To calculate a mole ratio, you need to divide the number of moles of one substance by the number of moles of the other substance.

Q: How do I calculate a mass ratio?

A: To calculate a mass ratio, you need to divide the mass of one substance by the mass of the other substance.

Q: What is the significance of calculating mole ratios and mass ratios?

A: Calculating mole ratios and mass ratios is important because it helps you predict the yield of a product and design experiments to produce a specific amount of a substance.

Q: Can you give an example of how to calculate a mole ratio and a mass ratio?

A: Yes, let's consider the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O):

2H2+O2→2H2O{ 2H_2 + O_2 \rightarrow 2H_2O }

To calculate the mole ratio of H2 to O2, we can divide the number of moles of H2 by the number of moles of O2:

nH2nO2=2 mol1 mol=2{ \frac{n_{H_2}}{n_{O_2}} = \frac{2 \, mol}{1 \, mol} = 2 }

To calculate the mass ratio of H2 to O2, we can divide the mass of H2 by the mass of O2:

mH2mO2=2 g1 g=2{ \frac{m_{H_2}}{m_{O_2}} = \frac{2 \, g}{1 \, g} = 2 }

Therefore, the mole ratio of H2 to O2 is 2:1, and the mass ratio of H2 to O2 is also 2:1.