Solution Stoichiometry: Videos & Practice Problems

Stoichiometry is a crucial concept in chemistry that involves using a balanced chemical equation to determine the relationship between the quantities of reactants and products. When incorporating solutions into stoichiometry, volume and molarity become important factors. For instance, if given the volume and molarity of water in a reaction with sodium to produce hydrogen gas and sodium hydroxide, one can calculate the grams of hydrogen gas produced. The process involves converting volume to liters if necessary, using molarity to find the moles of the given substance, and then using the mole ratio from the balanced equation to find the moles of the desired product. Finally, the moles of the product can be converted into grams using the molar mass. This approach is an extension of stoichiometry that now accounts for the volume and molarity of solutions in the calculations.

Solution Stoichiometry

Solution stoichiometry focuses on calculations involving the concentration and volume of solutions, integrating these concepts into traditional stoichiometric methods. In this context, a stoichiometric chart is adapted to accommodate scenarios where volume and molarity are provided. This chart allows for the conversion of a known quantity of one compound into an unknown quantity of another, facilitating the calculation process.

For instance, consider a balanced chemical equation where 2 moles of solid sodium react with 2 moles of liquid water to produce 1 mole of hydrogen gas and 2 moles of sodium hydroxide in aqueous form. If we are given 38.74 milliliters of a 0.275 molar solution of water, we can determine the mass of hydrogen gas produced.

To begin, it is essential to convert the volume of water from milliliters to liters, as the relationship between moles, volume, and molarity is defined by the equation:

moles = liters × molarity

By converting 38.74 mL to liters (0.03874 L) and multiplying by the molarity (0.275 mol/L), we can find the moles of water:

moles of H₂O = 0.03874 L × 0.275 mol/L = 0.01065 mol

Next, using the stoichiometric coefficients from the balanced equation, we perform a mole-to-mole conversion to find the moles of hydrogen gas produced. The balanced equation indicates that 2 moles of water yield 1 mole of hydrogen gas, leading to the following calculation:

moles of H₂ = 0.01065 mol H₂O × (1 mol H₂ / 2 mol H₂O) = 0.005325 mol H₂

Finally, to convert moles of hydrogen gas to grams, we use the molar mass of hydrogen (approximately 2.02 g/mol):

grams of H₂ = 0.005325 mol × 2.02 g/mol = 0.01076 g

This example illustrates how solution stoichiometry builds upon traditional stoichiometric principles by incorporating volume and molarity into the calculations. Understanding these relationships is crucial for accurately determining the quantities of reactants and products in chemical reactions.

Solution Stoichiometry Example

In this stoichiometry problem, we are tasked with determining the number of moles of hydrogen gas produced from a reaction involving water and excess sodium. The process begins with understanding the balanced chemical equation, which allows us to relate the quantities of reactants and products through stoichiometric coefficients.

To solve the problem, we first convert the given volume of water from milliliters to liters. Since there are 1000 milliliters in a liter, we can express 38.74 milliliters as:

38.74 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.03874 \, \text{L}

Next, we apply the molarity formula, which states that moles of solute equals the volume of solution in liters multiplied by the molarity:

moles = \text{liters} \times \text{molarity}

Given that the molarity of water is 0.275 M, we can calculate the moles of water:

moles \, \text{of water} = 0.03874 \, \text{L} \times 0.275 \, \text{mol/L} = 0.01065 \, \text{mol}

Now, we perform a mole-to-mole conversion using the coefficients from the balanced equation. The equation indicates that 2 moles of water produce 1 mole of hydrogen gas (H₂). Therefore, we set up the conversion as follows:

moles \, \text{of } H_2 = 0.01065 \, \text{mol} \times \frac{1 \, \text{mol} \, H_2}{2 \, \text{mol} \, \text{water}} = 0.005325 \, \text{mol} \, H_2

Thus, the final result indicates that 5.33 x 10^-3 moles of hydrogen gas are produced from the reaction. This calculation illustrates the application of stoichiometry in determining the amounts of reactants and products in a chemical reaction, emphasizing the importance of balanced equations and mole conversions in the process.

How many milliliters of 0.325 M HCl are needed to react with 16.2 g of magnesium metal? 
2 HCl (aq) + Mg (s) → MgCl₂ + H₂ (g)

• A. 33 mL
• B. 742 mL
• C. 872 mL
• D. 3321 mL
• E. 4100 mL

Answer: (E)

What is the molar concentration of a hydrobromic acid solution if it takes 34.12 mL of HBr to completely neutralize 82.56 mL of 0.156 M Ca(OH)­₂? 
2 HBr (aq) + Ca(OH)₂ (aq)  → CaBr₂ (aq) + 2 H₂O (l)

• A. 0.326 M
• B. 0.755 M
• C. 0.811 M
• D. 1.38 M
• E. 1.85 M

Answer: (B)

Consider the following balanced chemical equation: 
H₂O+ 2 MnO₄^– + 3 SO₃^2-  → 2 MnO₂ + 3 SO₄^2-+ 2 OH^–
How many grams of MnO₂ (MW:86.94 g/mol) will be created when 25.0 mL of 0.120 M MnO₄^– (MW:118.90 g/mol) reacts with 32.0 mL of 0.140 M SO₃^2- (MW:80.07 g/mol).  

• A. 0.073 g
• B. 0.089 g
• C. 0.19 g
• D. 0.26 g
• E. 0.67 g

Answer: (D)

Here’s what students ask on this topic:

What is solution stoichiometry and how is it different from regular stoichiometry?

Solution stoichiometry involves stoichiometric calculations in solutions that include volume and molarity, unlike regular stoichiometry which typically deals with masses and moles. In solution stoichiometry, you often start with the volume and molarity of a solution to find the moles of a substance. This is then used in conjunction with a balanced chemical equation to determine the moles of another substance. Finally, these moles can be converted to grams or other units as needed. This approach extends regular stoichiometry by incorporating the concentration of solutions into the calculations.

How do you calculate the grams of a product formed in a solution stoichiometry problem?

To calculate the grams of a product in a solution stoichiometry problem, follow these steps: 1) Convert the given volume to liters if necessary. 2) Use the molarity to find the moles of the given substance using the formula: $\mathrm{moles}=\mathrm{volume}\times \mathrm{molarity}$. 3) Use the mole ratio from the balanced chemical equation to find the moles of the desired product. 4) Convert the moles of the product to grams using its molar mass. This process ensures accurate stoichiometric calculations involving solutions.

What is the role of molarity in solution stoichiometry?

Molarity (M) is a measure of the concentration of a solution, defined as moles of solute per liter of solution. In solution stoichiometry, molarity is crucial because it allows you to convert between the volume of a solution and the moles of solute it contains. This conversion is essential for using the balanced chemical equation to determine the quantities of reactants and products. The formula used is: $\mathrm{moles}=\mathrm{volume}\times \mathrm{molarity}$.

How do you use a balanced chemical equation in solution stoichiometry?

In solution stoichiometry, a balanced chemical equation is used to determine the mole ratio between reactants and products. This ratio is essential for converting the moles of a given substance (found using volume and molarity) to the moles of the desired product. For example, if the balanced equation is $2\mathrm{Na}+2H2O\to H2+2\mathrm{NaOH}$, and you know the moles of $H2O$, you can use the ratio to find the moles of $H2$.

What is the significance of the mole-to-mole comparison in solution stoichiometry?

The mole-to-mole comparison in solution stoichiometry is significant because it allows you to relate the quantities of different substances in a chemical reaction. Using the coefficients from the balanced chemical equation, you can determine how many moles of one substance react with or produce a certain number of moles of another substance. This comparison is essential for converting the moles of a given reactant (found using volume and molarity) to the moles of the desired product, which can then be further converted to grams or other units.