Ideal Gas Law & Solutions (College Board AP® Chemistry)

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Martín

Written by: Martín

Reviewed by: Stewart Hird

Calculations of Gases & Solutions

Calculations with Ideal Gases

  • Gases that participate in chemical reactions can be used to perform mole calculations

  • It must be assumed that all the gasses behave as ideal gases

    • Therefore, the ideal gas equation must be used depending on the context of the problem

  • The ideal gas equation connects the pressure (P), volume (V) and temperature (T) of an ideal gas with the amount of moles (n)

    • The equation is shown below:

 PV = nRT

  • The gas constant (R) can adopt different values depending on the units used for pressure, volume and temperature

    • R = 8.314 J mol-1 K-1 = 0.08206 L atm mol-1 K-1 = 62.36 L torr mol-1 K-1

  • These values will be shown in the Equations and Constants section at the beginning of your AP Chemistry examination

Worked Example

The reaction of calcium carbonate (CaCO3) and hydrochloric acid (HCl) produces calcium chloride (CaCl2), water (H2O), and carbon dioxide gas (CO2). If the reaction takes place in a container at 1.02 atm and 27.0 ℃. How many liters of carbon dioxide are released when 50.0 g of sodium carbonate reacts with enough hydrochloric acid?

Answer:

  • Step 1: Write a balanced equation for the chemical reaction

CaCO3 + HCl → CaCl2 +H2O + CO2

Balance Cl

CaCO3 + 2HCl → CaCl2 +H2O + CO2

 

Left

Right

Ca

1

1

C

1

1

O

3

3

H

2

2

Cl

2

2

 The equation is balanced

  • Step 2: Analyze the statement and set up the steps that you are going to use to solve the problem. The initial quantity is always given by the statement

calculations-with-gases-worked-example
  • Step 3: Calculate the moles of calcium carbonate using its mass and molar mass

M = relative molecular mass of CaCO3

M = (1 × mass of Ca) + (1 × mass of C) + (3 × mass of O)

M = (1 × 40.08) + (1 × 12.01) + (3 × 16.00)

M = 100.09 g mol-1

n = m/M

n = 50.0 g/ 100.09 g mol-1

n =  0.49955 mol of CaCO3

  • Step 4: Calculate the moles of carbon dioxide using the ratio from the chemical equation

begin mathsize 14px style 0.49955 space mol space of space CaCO subscript 3 space cross times fraction numerator 1 space mol space of space CO subscript 2 over denominator 1 space mol space of space CaCO subscript 3 end fraction equals space 0.49955 space mol space of space CO subscript 2 end style

Since the ratio is 1:1 the moles of carbon dioxide must be the same

  • Step 5: Calculate the liters of carbon dioxide using the ideal gas equation

PV = nRT

Since the answer must be in liters, R =  0.08206 L atm mol-1 K-1 

The pressure and temperature are given by the statement: 1.02 atm and 27.0 ℃ respectively

When working with the ideal gas equation, temperature must be in Kelvin. Therefore,

K = °C + 273

K = 27.0 + 273

K = 300 K

Once, temperature is in Kelvin and pressure in atm. The ideal gas equation can be applied

PV = nRT

Rearranging the equation,

straight V space equals space nRT over straight P

Replacing the variables by the quantities given by the statement,

straight V space equals space fraction numerator left parenthesis 0.49955 space mol right parenthesis left parenthesis 0.08206 space straight L space atm space mol to the power of negative 1 space end exponent straight K to the power of negative 1 end exponent right parenthesis left parenthesis 300 straight K right parenthesis over denominator 1.02 space atm end fraction

V = 12.057 L

  • Step 6: Write the answer with the appropriate number of significant figures.


    Since all the quantities from the statement are given with 3 significant figures. The answer must be written with 3 significant figures


    The volume of carbon dioxide produced is 12.1 L

Calculation with Solutions

  • There are a lot of chemical reactions between solutions

    • A solution is a liquid homogeneous mixture that has two components: a solute (in small amounts) a solvent (in big amounts)

    • The solvent in most of the reaction is water

  • The amount of moles that are involved in the reaction depends on the concentration of the solution and its volume

  • Molarity (M) is used to quantify the concentration, and it connects the moles of solute (n) and the volume of solution in liters (V)

straight M space equals space fraction numerator straight n space of space solute over denominator straight V space of space solution space in space liters end fraction

  • Remember that L = dm3

Worked Example

How much L of 0.100 M KCl solution will react completely with 0.200 L of a 0.200 M Pb(NO3)2?

2KCl (aq) + Pb(NO3)(aq) → PbCl2 (aq) + 2KNO3 (aq)

Answer:

  • Step 1: Analyze the statement and set up the steps that you are going to use to solve the problem. The initial quantity is always given by the statement. In calculations with solutions, start always with the reactant or product that you have the most information

Since, volume and molarity are given for Pb(NO3)2, it must be the starting point

calculations-with-solutions-worked-example
  • Step 2: Calculate the moles of Pb(NO3)2 using the molarity equation

straight M space equals space fraction numerator straight n space of space solute over denominator straight V space of space solution space in space liters end fraction

Rearranging the equation,

n of solute = M × V of solution in liters

Replacing the values,

n of solute = 0.200 M × 0.200 L

n of solute = 0.04 mol of Pb(NO3)2

  • Step 3: Calculate the moles of KCl that reacted using the ratio from the chemical equation

0.04 space mol space of space Pb left parenthesis NO subscript 3 right parenthesis subscript 2 space cross times space fraction numerator 2 space mol space of space KCl over denominator 1 space mol space of space Pb left parenthesis NO subscript 3 right parenthesis subscript 2 end fraction space equals space 0.08 space mol space of space KCl

Since the ratio is 1:2 the moles of KCl must be double

  • Step 4: Calculate the volume in L of KCl solution that are needed to react completely

straight M space equals space fraction numerator straight n space of space solute over denominator straight V space of space solution space in space liters end fraction

Rearranging the equation,

straight V space of space solution space in space liters space equals space fraction numerator straight n space of space solute over denominator straight M end fraction

Replacing the values,

straight V space of space solution space in space liters space equals space fraction numerator 0.08 space mol space of space KCl over denominator 0.100 space straight M end fraction

straight V space of space solution space in space liters space equals space 0.8 space straight L

  • Step 5: Write the answer with the appropriate number of significant figures


    Since all the quantities from the statement are given with 3 significant figures. The answer must be written with 3 significant figures


    The volume of KCl solution required to react completely with the Pb(NO3)2 solution  is 0.800 L

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Martín

Author: Martín

Expertise: Chemistry Content Creator

Martín, a dedicated chemistry teacher and tutor, excels in guiding students through IB, AP, and IGCSE Chemistry. As an IB Chemistry student, he came from hands-on preparation, focusing on practical exam techniques and rigorous practice. While at Universidad San Francisco de Quito, his academic journey sparked a passion for computational and physical chemistry. Martín specializes in chemistry, and he knows that SaveMyExams is the right place if he wants to have a positive impact all around the world.

Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Lead

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.