Consider the chemical reaction: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ How many liters of hydrogen gas are formed from the complete reaction of \(1.07 \mathrm{~mol}\) of \(\mathrm{C} ?\) Assume that the hydrogen gas is collected at \(1.0 \mathrm{~atm}\) and \(315 \mathrm{~K}\).

Short Answer

Expert verified
28.06 liters of hydrogen gas are formed from the complete reaction of 1.07 mol of Carbon.

Step by step solution

01

Write the Balanced Chemical Equation

First, ensure that the given chemical reaction is balanced. The given reaction is balanced as written: \[\mathrm{C}(s) + \mathrm{H}_2\mathrm{O}(g) \longrightarrow \mathrm{CO}(g) + \mathrm{H}_{2}(g)\]
02

Determine Moles of Hydrogen Gas Produced

From the stoichiometry of the balanced equation, 1 mole of Carbon reacts with 1 mole of water to produce 1 mole of Hydrogen gas. Therefore, if 1.07 mol of Carbon reacts completely, it will produce 1.07 mol of Hydrogen gas.
03

Use Ideal Gas Law to Find Volume of Hydrogen Gas

The ideal gas law, \(PV = nRT\), can be rearranged to solve for volume \(V\) where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. Here, \(P = 1.0 \text{ atm}\), \(n = 1.07 \text{ mol}\), \(R = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}\) and \(T = 315 \text{ K}\). Plugging in the values, we find the volume of hydrogen gas.
04

Calculate the Volume of Hydrogen Gas

Using the rearranged ideal gas law \(V = \frac{nRT}{P}\), we get: \[V = \frac{(1.07 \text{ mol})(0.0821 \text{ L atm mol}^{-1}\text{K}^{-1})(315 \text{ K})}{1.0 \text{ atm}}\]Now calculate the value to get the volume in liters.
05

Perform the Calculation

Substitute the known values into the equation and calculate the volume: \[V = \frac{(1.07 \text{ mol})(0.0821 \text{ L atm}/\text{mol K})(315 \text{ K})}{1.0 \text{ atm}}\]Solve for V to get the volume of hydrogen gas produced.

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Key Concepts

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

Balanced Chemical Equation
Understanding chemical reactions begins with a balanced chemical equation. This equation represents the law of conservation of mass, where the number of atoms of each element remains the same before and after the reaction. It's crucial because it allows us to predict how much of each substance will react and what will be produced. For instance, in the reaction \[ \mathrm{C}(s) + \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g) + \mathrm{H}_{2}(g), \] each element has an equal amount of atoms on both the reactant and product sides, indicating a balanced equation. Balancing chemical equations requires making sure that the number of atoms of each element on both sides of the equation is equal, often by adjusting coefficients in front of the chemical formulas.

For educational clarity, visual representation of molecules before and after the reaction could aid understanding. In this case, imagining a single carbon atom reacting with a water molecule to produce a carbon monoxide molecule and a molecule of hydrogen gas can help visualize the stoichiometric relationships.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics, which describes the behavior of an 'ideal' gas under various conditions. The law is typically stated as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. This law helps us to calculate one property of a gas if the other three are known. For example, we used the ideal gas law in the exercise to find the volume (\( V \)) of hydrogen gas formed from a given number of moles (\( n \)), at a specific temperature (\( T \)) and pressure (\( P \)).

It is essential to express all quantities in the correct units and to use Kelvin for temperature. It can be helpful to explain that 0 degrees Celsius corresponds to 273.15 Kelvin, to ease the students' conversion. In real-life applications, the ideal gas law assumes no intermolecular forces and that particles occupy no space, which is nearly true for gases at high temperatures and low pressures.
Molar Volume of a Gas
The molar volume of a gas is defined as the volume occupied by one mole of the gas at a given temperature and pressure. For an ideal gas at standard temperature and pressure (0 degrees Celsius and 1 atmosphere), the molar volume is approximately 22.4 liters. It's essential to understand that real gases don't always behave ideally; this standard molar volume is an approximation. However, it provides a useful base for comparison and calculation in stoichiometry problems.

When dealing with the molar volume at conditions other than standard temperature and pressure, we can rely on the ideal gas law to calculate the actual volume occupied by the gas. This is what we did in the step-by-step solution to determine how many liters of hydrogen gas would be produced under the given conditions.
Mole-to-Mole Conversion
Mole-to-mole conversion is a crucial aspect of stoichiometry, which helps to understand the relationships between the reactants and products in a chemical reaction. According to the stoichiometry of a reaction, you can determine how many moles of one substance will react with or produce a certain number of moles of another substance. In the exercise, we used mole-to-mole conversion to infer that 1.07 moles of carbon would produce 1.07 moles of hydrogen gas, based on the balanced chemical equation.

For students, practicing mole-to-mole conversions is vital. They should get comfortable with using the coefficients in a balanced equation to translate moles of one substance into moles of another. This defines the exact proportions in which chemicals react. Visual aids, such as mole maps or flowcharts, can be exceptionally beneficial to guide students through these conversions systematically.

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Most popular questions from this chapter

Consider the reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{~s}) $$ (a) If \(285.5 \mathrm{~mL}\) of \(\mathrm{SO}_{2}\) is allowed to react with \(158.9 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) (both measured at \(\mathrm{STP}\) ), what are the limiting reactant and the theoretical yield of \(\mathrm{SO}_{3}\) ? (b) If \(2.805 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) is collected (measured at \(\mathrm{STP}\) ), what is the percent yield for the reaction?

What is standard temperature and pressure (STP)? What is the molar volume of a gas at STP?

Perform each conversion. (a) \(1.06 \mathrm{~atm}\) to millimeters of mercury (b) \(95,422 \mathrm{~Pa}\) to millimeters of mercury (c) \(22.3\) psi to torr (d) \(35.78\) in. Hg to millimeters of mercury

Consider the reaction for the production of \(\mathrm{NO}_{2}\) from \(\mathrm{NO}\) : $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ (a) If \(84.8 \mathrm{~L}\) of \(\mathrm{O}_{2}(g)\), measured at \(35^{\circ} \mathrm{C}\) and \(632 \mathrm{~mm} \mathrm{Hg}\), is allowed to react with \(158.2 \mathrm{~g}\) of \(\mathrm{NO}\), find the limiting reagent. (b) If \(97.3 \mathrm{~L}\) of \(\mathrm{NO}_{2}\) forms, measured at \(35^{\circ} \mathrm{C}\) and \(632 \mathrm{~mm}\) \(\mathrm{Hg}\), what is the percent yield?

A cylinder contains \(11.8 \mathrm{~L}\) of air at a total pressure of \(43.2\) psi and a temperature of \(25^{\circ} \mathrm{C}\). How many moles of gas does the cylinder contain? (Hint: You must convert each quantity into the correct units \((\mathrm{L}, \mathrm{atm}, \mathrm{mol}\), and \(\mathrm{K})\) before substituting into the ideal gas law.)

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