Question

Wow many grams of water can be produced from the combination of 25 grams of hydrogen and 225 grams of oxygen? How much of which element will be lcft over?

Short answer

225 grams of water are produced, and 24.96 grams of oxygen are left over.

Step-by-step solution

Balanced Chemical Equation

First, write the balanced chemical equation for the formation of water from hydrogen and oxygen: \[ 2H_2 + O_2 \rightarrow 2H_2O \]This equation indicates that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

Molar Mass Calculation

Calculate the molar masses of the substances involved: - The molar mass of hydrogen (\(H_2\)) is approximately 2 g/mol.- The molar mass of oxygen (\(O_2\)) is approximately 32 g/mol.- The molar mass of water (\(H_2O\)) is approximately 18 g/mol.

Determine Moles of Reactants

Calculate the number of moles of each reactant:For hydrogen:\[ \text{Moles of } H_2 = \frac{25 \text{ grams}}{2 \text{ g/mol}} = 12.5 \text{ moles} \]For oxygen:\[ \text{Moles of } O_2 = \frac{225 \text{ grams}}{32 \text{ g/mol}} = 7.03 \text{ moles} \]

Identify the Limiting Reactant

According to the balanced equation, 7.03 moles of \(O_2\) would require 14.06 moles of \(H_2\) (since 1 mole of \(O_2\) requires 2 moles of \(H_2\)). However, we only have 12.5 moles of \(H_2\). Thus, \(H_2\) is the limiting reactant.

Calculate Water Production

Using the amount of the limiting reactant \(H_2\):\[ 12.5 \text{ moles of } H_2 \rightarrow 12.5 \text{ moles of } H_2O \]Calculate the mass of water produced:\[ \text{Mass of } H_2O = 12.5 \text{ moles} \times 18 \text{ g/mol} = 225 \text{ grams} \]

Calculate Leftover Oxygen

Since \(H_2\) is the limiting reactant, some \(O_2\) will be left over. Calculate the moles of \(O_2\) reacted:\[ \text{Moles of } O_2 \text{ used} = \frac{12.5}{2} = 6.25 \text{ moles} \]Calculate the mass of leftover \(O_2\):\[ \text{Leftover } O_2 = (7.03 - 6.25) \text{ moles} \times 32 \text{ g/mol} = 24.96 \text{ grams} \]

Key concepts

Balanced Chemical Equations

In chemistry, a balanced chemical equation is crucial for understanding the stoichiometry of a reaction. It shows how atoms rearrange during the reaction while obeying the Law of Conservation of Mass. This law states that matter is neither created nor destroyed, only transformed.
To illustrate this, consider the formation of water from hydrogen and oxygen:

• The balanced equation is: \[ 2H_2 + O_2 \rightarrow 2H_2O \] .
• Here, 2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.
• The coefficients in front of each molecule represent the moles needed to balance the atoms on both sides of the equation.

Balancing equations ensures you have the correct proportions of reactants and products, a necessary step in calculating the yield of a reaction.

Limiting Reactants

The concept of the limiting reactant is essential in chemical reactions. The limiting reactant is the substance that is completely consumed when the chemical reaction is complete.
This reactant limits the amount of product formed. In the reaction between hydrogen and oxygen:

• We determined that \(H_2\) is the limiting reactant because it runs out before \(O_2\) does.
• The balanced equation \(2H_2 + O_2 \rightarrow 2H_2O \) tells us you need twice as much hydrogen as oxygen to react completely.
• In this exercise, there are only 12.5 moles of \(H_2\) available, but 7.03 moles of \(O_2\) are present, where 14.06 moles of \(H_2\) would be required for complete reaction.
• This means that not all \(O_2\) can react due to the insufficient amount of \(H_2\). Therefore, \(H_2\) determines the quantity of water produced.

Knowing which reactant is limiting helps predict the exact amount of product that can be formed and any excess reactants leftover.

Molar Mass Calculations

Molar mass calculations are necessary for converting between grams and moles, a common task in stoichiometry. Molar mass, often expressed in grams per mole (g/mol), is the mass of one mole of a given substance. Here’s how it works with hydrogen (\(H_2\)), oxygen (\(O_2\)), and water (\(H_2O\)):

• The molar mass of \(H_2\) is approximately 2 g/mol because each hydrogen atom has a molar mass of about 1 g/mol, and there are two hydrogen atoms in a molecule of \(H_2\).
• The molar mass of \(O_2\) is about 32 g/mol, with each oxygen atom having a molar mass of 16 g/mol.
• For \(H_2O\), the total molar mass is around 18 g/mol. This value comes from the sum of 2 g/mol for two hydrogen atoms and 16 g/mol for one oxygen atom.

These calculations are essential for determining how much of each substance is involved in the reaction:

• Convert the mass of a reactant into moles to find out how many molecules are available to react.
• Once the moles are calculated, they can be used in the balanced equation to identify limiting reactants and calculate theoretical yields.
• In this problem, they help compare the amount of reactants to ensure proper amounts for complete reaction and find leftover substances.

Understanding molar mass helps bridge the gap between the microscopic world of atoms and the macroscopic world we can measure and observe.