Stoichiometry and the Balanced Equation

Oct 13, 2025 by orovajewels.com 1 views

Introduction

In this problem, we are tasked with determining the volume of 3.0 M HCl (aq) required to react with 0.50 g of CaCO3 (s) given the reaction: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). To solve this problem, we need to use the concept of stoichiometry, which is the study of the quantitative relationships between reactants and products in chemical reactions.

Stoichiometry and the Balanced Equation

The balanced equation for the reaction between CaCO3 and HCl is:

CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

From the balanced equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl. We can use this information to determine the number of moles of HCl required to react with 0.50 g of CaCO3.

Calculating the Number of Moles of CaCO3

To calculate the number of moles of CaCO3, we need to know its molar mass. The molar mass of CaCO3 is 100.09 g/mol. We can use this value to calculate the number of moles of CaCO3 as follows:

moles CaCO3 = mass CaCO3 / molar mass CaCO3 moles CaCO3 = 0.50 g / 100.09 g/mol moles CaCO3 = 0.005 mol

Calculating the Number of Moles of HCl Required

Since 1 mole of CaCO3 reacts with 2 moles of HCl, we can calculate the number of moles of HCl required as follows:

moles HCl = 2 x moles CaCO3 moles HCl = 2 x 0.005 mol moles HCl = 0.01 mol

Calculating the Volume of HCl Required

Now that we know the number of moles of HCl required, we can use the molarity of the HCl solution to calculate the volume of HCl required. The molarity of the HCl solution is 3.0 M, which means that 1 liter of the solution contains 3.0 moles of HCl. We can use this value to calculate the volume of HCl required as follows:

volume HCl = moles HCl / molarity HCl volume HCl = 0.01 mol / 3.0 M volume HCl = 0.00333 L

Converting the Volume from Liters to Milliliters

Since the problem asks for the volume of HCl in milliliters, we need to convert the volume from liters to milliliters. There are 1000 milliliters in 1 liter, so we can convert the volume as follows:

volume HCl (mL) = volume HCl (L) x 1000 volume HCl (mL) = 0.00333 L x 1000 volume HCl (mL) = 3.33 mL

The final answer is: 3.33

Introduction

In our previous article, we discussed how to determine the volume of 3.0 M HCl (aq) required to react with 0.50 g of CaCO3 (s) given the reaction: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). In this article, we will answer some frequently asked questions related to this problem.

Q: What is the concept of stoichiometry in chemistry?

A: Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. It helps us to determine the amount of reactants required to produce a certain amount of products.

Q: How do we calculate the number of moles of CaCO3?

A: To calculate the number of moles of CaCO3, we need to know its molar mass. The molar mass of CaCO3 is 100.09 g/mol. We can use this value to calculate the number of moles of CaCO3 as follows:

moles CaCO3 = mass CaCO3 / molar mass CaCO3 moles CaCO3 = 0.50 g / 100.09 g/mol moles CaCO3 = 0.005 mol

Q: How do we calculate the number of moles of HCl required?

A: Since 1 mole of CaCO3 reacts with 2 moles of HCl, we can calculate the number of moles of HCl required as follows:

moles HCl = 2 x moles CaCO3 moles HCl = 2 x 0.005 mol moles HCl = 0.01 mol

Q: How do we calculate the volume of HCl required?

A: Now that we know the number of moles of HCl required, we can use the molarity of the HCl solution to calculate the volume of HCl required. The molarity of the HCl solution is 3.0 M, which means that 1 liter of the solution contains 3.0 moles of HCl. We can use this value to calculate the volume of HCl required as follows:

volume HCl = moles HCl / molarity HCl volume HCl = 0.01 mol / 3.0 M volume HCl = 0.00333 L

Q: How do we convert the volume from liters to milliliters?

A: Since the problem asks for the volume of HCl in milliliters, we need to convert the volume from liters to milliliters. There are 1000 milliliters in 1 liter, so we can convert the volume as follows:

volume HCl (mL) = volume HCl (L) x 1000 volume HCl (mL) = 0.00333 L x 1000 volume HCl (mL) = 3.33 mL

Q: What is the final answer?

A: The final answer is 3.33 mL.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

Not using the correct molar mass of CaCO3
Not using the correct molarity of the HCl solution
Not converting the volume from liters to milliliters
Not using the correct stoichiometry of the reaction

Q: How can I practice solving this type of problem?

A: You can practice solving this type of problem by trying different values of mass of CaCO3 and molarity of the HCl solution. You can also try solving similar problems with different reactants and products.

Q: What are some real-world applications of this type of problem?

A: This type of problem has many real-world applications in fields such as chemistry, chemical engineering, and environmental science. For example, it can be used to determine the amount of chemicals required to clean up a spill or to design a chemical reactor.

The final answer is: 3.33