Limiting Reactant And Water Mass Calculation

Oct 30, 2025 by ADMIN 45 views

Hey there, chemistry enthusiasts! Let's dive into a classic problem involving limiting reactants and the calculation of product mass. This is a fundamental concept in stoichiometry, and mastering it will set you up for success in many chemistry topics. We'll break down the steps, making it easy to understand, even if you're just starting out. Our task is to determine the limiting reactant and calculate the mass of water ( $H_2O$ ) produced, given an unbalanced chemical equation and the masses of the reactants. Buckle up, and let's get started!

First, let's understand what a limiting reactant is. Imagine you're making sandwiches. You have 10 slices of bread and 5 slices of cheese. You need 2 slices of bread for each sandwich and 1 slice of cheese. You can only make 2 sandwiches because you run out of cheese first. The cheese is your limiting factor! The limiting reactant in a chemical reaction is the reactant that is completely consumed first, thus limiting the amount of product that can be formed. The other reactants are present in excess.

Unveiling the Chemical Equation

Before we can tackle the problem, we need a chemical equation. Let's assume the unbalanced chemical equation is as follows:

$Mg(s) + O_2(g) → MgO(s)$

This equation represents the reaction of magnesium ( $Mg$ ) with oxygen ( $O_2$ ) to produce magnesium oxide ( $MgO$ ). Now, this equation is unbalanced. To make sure we're on the right track, let's balance this equation. Balancing an equation ensures that the number of atoms of each element is the same on both sides of the equation, following the law of conservation of mass. To balance it, we get:

$2Mg(s) + O_2(g) → 2MgO(s)$

This balanced equation tells us that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide. This is the foundation upon which we'll build our calculations, so it's super important to get it right. Also, we will use 1.70 g of each reactant.

Step-by-Step Calculation of the Limiting Reactant

Now, let's figure out which reactant is limiting. We'll be using the masses provided (1.70 g of each reactant) and the balanced chemical equation. Here's a step-by-step approach:

Convert grams to moles: We need to convert the mass of each reactant into moles. To do this, we'll use the molar mass of each substance. The molar mass is the mass of one mole of a substance and is found on the periodic table.
- For Magnesium ( $Mg$ ): The molar mass is approximately 24.31 g/mol.
- For Oxygen ( $O_2$ ): The molar mass is approximately 32.00 g/mol.
- Moles of Magnesium: $(1.70 g Mg) / (24.31 g/mol) = 0.070 mol Mg$
- Moles of Oxygen: $(1.70 g O_2) / (32.00 g/mol) = 0.053 mol O_2$
Determine the mole ratio from the balanced equation: The balanced equation $2Mg(s) + O_2(g) → 2MgO(s)$ tells us the mole ratio between the reactants. It tells us that 2 moles of magnesium react with 1 mole of oxygen.
Calculate the amount of one reactant needed to react with the other: We can use the mole ratio to figure out how much of one reactant is needed to react completely with the other. Let's start by calculating how much oxygen is needed to react with all the magnesium.
- Moles of $O_2$ needed = $(0.070 mol Mg) * (1 mol O_2 / 2 mol Mg) = 0.035 mol O_2$
Compare the actual moles to the required moles: Now, compare the amount of oxygen we have to the amount of oxygen we need. We have 0.053 mol of $O_2$ , but we only need 0.035 mol of $O_2$ to react with all the magnesium. Since we have more oxygen than we need, oxygen is in excess, and magnesium is the limiting reactant.

Therefore, Magnesium (Mg) is the limiting reactant.

Calculating the Mass of $H_2O$

Now that we've identified the limiting reactant, we can calculate the mass of water ( $H_2O$ ) produced. We'll use the moles of the limiting reactant (magnesium) and the stoichiometry of the balanced equation. However, the provided equation is not the equation for the formation of water. So let's use the following equation to show the concept.

$2H_2(g) + O_2(g) ightarrow 2H_2O(l)$

Convert grams to moles: We need to convert the mass of each reactant into moles. To do this, we'll use the molar mass of each substance. The molar mass is the mass of one mole of a substance and is found on the periodic table.
- For Hydrogen ( $H_2$ ): The molar mass is approximately 2.02 g/mol.
- For Oxygen ( $O_2$ ): The molar mass is approximately 32.00 g/mol.
- Moles of Hydrogen: $(1.70 g H_2) / (2.02 g/mol) = 0.84 mol H_2$
- Moles of Oxygen: $(1.70 g O_2) / (32.00 g/mol) = 0.053 mol O_2$
Determine the mole ratio from the balanced equation: The balanced equation $2H_2(g) + O_2(g) ightarrow 2H_2O(l)$ tells us the mole ratio between the reactants. It tells us that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Determine the limiting reactant: We can use the mole ratio to figure out how much of one reactant is needed to react completely with the other. Let's start by calculating how much oxygen is needed to react with all the hydrogen.
- Moles of $O_2$ needed = $(0.84 mol H_2) * (1 mol O_2 / 2 mol H_2) = 0.42 mol O_2$
- Moles of $H_2$ needed = $(0.053 mol O_2) * (2 mol H_2 / 1 mol O_2) = 0.106 mol H_2$
Since we need 0.42 moles of $O_2$ but only have 0.053 moles of $O_2$ , so $O_2$ is the limiting reactant. Also, we need 0.106 moles of $H_2$ but we have 0.84 moles of $H_2$ , so $H_2$ is the excess reactant.
Calculate the moles of $H_2O$ produced using the limiting reactant: From the balanced equation, we know that 1 mole of $O_2$ produces 2 moles of $H_2O$ . Since $O_2$ is the limiting reactant, we use the moles of $O_2$ to calculate the moles of $H_2O$ produced.
- Moles of $H_2O$ produced = $(0.053 mol O_2) * (2 mol H_2O / 1 mol O_2) = 0.106 mol H_2O$
Convert moles of $H_2O$ to grams: Now, we'll convert the moles of $H_2O$ to grams using the molar mass of water (approximately 18.01 g/mol).
- Mass of $H_2O$ produced = $(0.106 mol H_2O) * (18.01 g/mol) = 1.91 g H_2O$

Therefore, we expect to produce approximately 1.91 g of water ( $H_2O$ ), assuming the limiting reactant is completely consumed.

Conclusion

And there you have it! We've successfully determined the limiting reactant and calculated the mass of water produced. Understanding limiting reactants is a crucial step in understanding stoichiometry and predicting the outcome of chemical reactions. Keep practicing, and you'll become a pro in no time! Remember, the key is to balance the equation, convert to moles, use the mole ratio, and work with the limiting reactant. Keep up the awesome work, and happy experimenting!