Solution Stoichiometry: Videos & Practice Problems

Solution stoichiometry involves calculations using volume and molarity to determine unknown quantities in chemical reactions. For example, using a balanced equation, one can convert the volume of a solution (in mL) and its molarity to moles, applying the relationship: ${\mathrm{moles}}^{=}\mathrm{liters}\mathrm{molarity}$. This process allows for mole-to-mole comparisons to find unknowns, such as grams of hydrogen gas produced in a reaction. Understanding this concept is essential for mastering stoichiometric calculations in chemistry.

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 determination of an unknown quantity of a compound based on a known quantity of another.

For instance, consider a balanced chemical equation where 2 moles of sodium (Na) react with 2 moles of water (H₂O) to produce 1 mole of hydrogen gas (H₂) and 2 moles of sodium hydroxide (NaOH). If we are given 38.74 milliliters of a 0.275 M solution of water, we can calculate the grams 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 M), 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 calculations. Understanding these concepts is crucial for solving various chemical problems, particularly in laboratory settings where solutions are frequently used.

Solution Stoichiometry Example 1

In this stoichiometry problem, we are tasked with determining the number of moles of hydrogen gas (H₂) produced from a reaction involving water (H₂O) and excess sodium (Na). The process begins with the understanding that stoichiometry allows us to use a balanced chemical equation to relate the quantities of reactants and products.

To solve the problem, we first convert the volume of water from milliliters to liters. Given that 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 (M). Here, the molarity of water is given as 0.275 M, meaning there are 0.275 moles of water per liter of solution. Thus, we can calculate the moles of water present:

\text{Moles of H}_2\text{O} = 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 chemical equation. The reaction between water and sodium produces hydrogen gas, and the balanced equation indicates that 2 moles of water yield 1 mole of hydrogen gas. Therefore, we set up the conversion as follows:

\text{Moles of H}_2 = 0.01065 \, \text{mol H}_2\text{O} \times \frac{1 \, \text{mol H}_2}{2 \, \text{mol H}_2\text{O}} = 0.005325 \, \text{mol H}_2

Finally, we can express the result in scientific notation:

0.005325 \, \text{mol H}_2 = 5.33 \times 10^{-3} \, \text{mol H}_2

This calculation shows that when 38.74 milliliters of 0.275 M water reacts with excess sodium, approximately 5.33 x 10^-3 moles of hydrogen gas are produced. Understanding these steps is crucial for solving similar stoichiometric problems in chemistry.

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 used in chemical reactions?

Solution stoichiometry involves using the volume and molarity of solutions to determine unknown quantities in chemical reactions. It extends traditional stoichiometry by incorporating the relationship between moles, volume, and molarity. For example, given a balanced chemical equation, you can convert the volume of a solution (in mL) and its molarity to moles using the equation:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

This allows for mole-to-mole comparisons to find unknowns, such as the grams of a product formed. Understanding this concept is crucial for mastering stoichiometric calculations in chemistry.

How do you convert volume and molarity to moles in solution stoichiometry?

To convert volume and molarity to moles in solution stoichiometry, use the equation:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

First, convert the volume from milliliters (mL) to liters (L) by dividing by 1000. Then, multiply the volume in liters by the molarity (M) of the solution. For example, if you have 38.74 mL of a 0.275 M solution, convert 38.74 mL to 0.03874 L and then multiply by 0.275 M to get the moles:

$0.038740.275=0.010654\mathrm{moles}$

How do you perform a mole-to-mole comparison in solution stoichiometry?

To perform a mole-to-mole comparison in solution stoichiometry, use the coefficients from the balanced chemical equation. For example, consider the reaction:

$2\mathrm{Na}\left(s\right)+2H2O\left(l\right)\to H2\left(g\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)$

If you know the moles of H₂O, you can find the moles of H₂ produced by using the ratio from the balanced equation. Here, 2 moles of H₂O produce 1 mole of H₂. So, if you have 0.010654 moles of H₂O, the moles of H₂ produced would be:

$\frac{0.010654}{2}=0.005327\mathrm{moles}H2$

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

To calculate the grams of a product formed in a solution stoichiometry problem, follow these steps:

1. Convert the volume and molarity of the reactant to moles using:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

2. Use the balanced chemical equation to perform a mole-to-mole comparison.

3. Convert the moles of the product to grams using its molar mass. For example, if you have 0.005327 moles of H₂ and the molar mass of H₂ is 2.02 g/mol, the mass in grams is:

$0.0053272.02=0.01076\mathrm{grams}$

What is the relationship between molarity, volume, and moles in solution stoichiometry?

The relationship between molarity, volume, and moles in solution stoichiometry is given by the equation:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

Molarity (M) is defined as the number of moles of solute per liter of solution. To find the moles of solute, multiply the volume of the solution (in liters) by its molarity. For example, if you have 0.5 L of a 1.5 M solution, the moles of solute are:

$0.51.5=0.75\mathrm{moles}$