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Two \(\mathrm{H}_{2}\) molecules can react with one \(\mathrm{O}_{2}\) molecule to produce two \(\mathrm{H}_{2} \mathrm{O}\) molecules. How many moles of hydrogen molecules are needed to react with one mole of oxygen molecules?

Short Answer

Expert verified
2 moles of hydrogen molecules are needed to react with one mole of oxygen molecules.

Step by step solution

01

Identify the reaction equation

The chemical reaction in this problem can be written as follows: \[ \text{2H}_{2} + \text{O}_{2} \rightarrow \text{2H}_{2}\text{O} \]
02

Relate the moles of reactants

From the equation, it is clear that 2 moles of hydrogen (\(\text{H}_{2}\)) are required to react with 1 mole of oxygen (\(\text{O}_{2}\)) to produce 2 moles of water (\(\text{H}_{2}\text{O}\)).
03

Determine the moles of hydrogen needed

Since 1 mole of oxygen (\(\text{O}_{2}\)) is given in the problem, we need to find the number of moles of hydrogen (\(\text{H}_{2}\)) needed to react with this oxygen. According to the equation, we already know that 2 moles of hydrogen are needed for 1 mole of oxygen. Therefore, the number of moles of hydrogen needed is: \[ \text{Moles of H}_{2} = 2 \times \text{Moles of O}_{2} \] For one mole of oxygen (\(\text{O}_{2}\)): \[ \text{Moles of H}_{2} = 2 \times \text{1 mole} \]
04

Calculate the moles of hydrogen

We can now calculate the number of moles of hydrogen needed to react with one mole of oxygen: \[ \text{Moles of H}_{2} = 2 \times \text{1 mole} = 2 \text{ moles} \] So, 2 moles of hydrogen molecules are needed to react with one mole of oxygen molecules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moles in Chemistry
Chemistry often deals with reactions between substances, where accurate measurement is vital. Here, the concept of a "mole" is essential. A mole is a way to quantify the "amount" of a chemical substance. It is similar to how a dozen refers to 12 items. Specifically, one mole of any substance contains Avogadro's number of entities, which is approximately \( 6.022 \times 10^{23} \) particles, atoms, ions, or molecules.

This large number comes from the atomic scale, where atoms and molecules are tiny and numerous. Chemists need to count these tiny particles in bulk to analyze chemical reactions.
  • It allows for predictions about the products and reactants involved in a chemical reaction.
  • All chemical equations are balanced on a mole basis, allowing for accurate ratio calculations.
Understanding moles helps in grasping concepts like molar mass and concentration, crucial for performing stoichiometric calculations in reactions.
Balanced Chemical Equations
In chemistry, reactions between substances must obey the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, chemical equations must be balanced to reflect this principle. A balanced chemical equation has the same number of each type of atom on both the reactant and product sides.

For example, consider the reaction:\[ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \]
  • There are 4 Hydrogen atoms and 2 Oxygen atoms on both sides.
  • It ensures mass and charge are conserved.
  • It helps in calculating the exact amounts of reactants needed for desired products.
Balancing equations can involve changing coefficients in front of compounds but never the subscripts within a formula, as this alters the compound's identity.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. The concept is rooted in the knowledge that "balanced" means that for a given chemical reaction, the amounts of reactants and products are related by whole number ratios derived from the coefficients in the balanced equation.

Using our example reaction, \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \), stoichiometry helps us determine that:
  • 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
  • These ratios guide chemists in practical tasks like scaling reactions up or down based on available materials.
By using stoichiometry, you can predict the outcomes of reactions and whether one reactant will limit the amount of product produced, known as the limiting reactant. This is crucial in both laboratory settings and industrial processes.

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Most popular questions from this chapter

The escape velocity from the Moon is much smaller than that from the Earth, only \(2.38 \mathrm{km} / \mathrm{s}\). At what temperature would hydrogen molecules (molar mass is equal to \(2.016 \mathrm{g} / \mathrm{mol}\) ) have a root-mean-square velocity \(v_{\mathrm{rms}}\) equal to the Moon's escape velocity?

a) Estimate the specific heat capacity of sodium from the Law of Dulong and Petit. The molar mass of sodium is 23.0 g/mol. (b) What is the percent error of your estimate from the known value, \(1230 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C} ?\)

(a) Hydrogen molecules (molar mass is equal to 2.016 g/mol) have \(v_{\text {rms }}\) equal to 193 m/s. What is the temperature? (b) Much of the gas near the Sun is atomic hydrogen (H rather than \(\mathrm{H}_{2}\) ). Its temperature would have to be \(1.5 \times 10^{7} \mathrm{K}\) for the rms speed \(v_{\mathrm{rms}}\) to equal the escape velocity from the Sun. What is that velocity?

Find the number of moles in \(2.00 \mathrm{L}\) of gas at \(35.0^{\circ} \mathrm{C}\) and under \(7.41 \times 10^{7} \mathrm{N} / \mathrm{m}^{2}\) of pressure.

In the chapter on fluid mechanics, Bernoulli's equation for the flow of incompressible fluids was explained in terms of changes affecting a small volume \(d V\) of fluid. Such volumes are a fundamental idea in the study of the flow of compressible fluids such as gases as well. For the equations of hydrodynamics to apply, the mean free path must be much less than the linear size of such a volume, \(a \approx d V^{1 / 3} .\) For air in the stratosphere at a temperature of 220 K and a pressure of 5.8 kPa, how big should a be for it to be 100 times the mean free path? Take the effective radius of air molecules to be \(1.88 \times 10^{-11} \mathrm{m},\) which is roughly correct for \(\mathrm{N}_{2}\)

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