The percent composition by mass of Na 2 CO 3 in the mixture has to be calculated. Concept Introduction: Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation. PV = nRT

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Answer 1

Question

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Conclusion

The percent composition by mass of $Na2CO3$ in the mixture was calculated.

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Answer 2

Question

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Conclusion

The percent composition by mass of $Na2CO3$ in the mixture was calculated.

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Answer 3

Question

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Conclusion

The percent composition by mass of $Na2CO3$ in the mixture was calculated.

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Answer 4

Question

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Conclusion

The percent composition by mass of $Na2CO3$ in the mixture was calculated.

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Answer 5

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Conclusion

The percent composition by mass of $Na2CO3$ in the mixture was calculated.

Answer 6

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Conclusion

The percent composition by mass of $Na2CO3$ in the mixture was calculated.

Answer 7

Chapter 5, Problem 5.115QP

Interpretation Introduction

Interpretation:

The percent composition by mass of $Na2CO3$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$PV = nRT$

Answer 8

Expert Solution & Answer

Answer to Problem 5.115QP

The percent composition by mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculate the moles of $CO2$ produced using the ideal gas equation.

$nCO2 =PVRT = (1.24 atm)(1.67 L)(0.0821L . atmK . mol) (26+ 273) K = 0.08436 mol CO2$

The moles of $CO2$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.

$0.08436 mol CO2$

Since there is a 1:1 mole ratio between $CO2$ and both reactants $(Na2CO3 and MgCO3),$ then 0.08436 mole of the mixture must have reacted.

$mol Na2CO3+ mol MgCO3 = 0.08436 mol$

$Let x be the mass of Na2CO3 in the mixture, then (7.63 - x) is the mass of MgCO3 in the mixture.$ $(x g Na2CO3 × 1 mol Na2CO3 106.0 g Na2CO3)+ [(7.63-x)g MgCO3 ×1 molMgCO3 84.32 g MgCO3 ]= 0.08436 mol$

$x = 2.527 g = mass of Na2CO3 in the mixture$

The mass of $Na2CO3$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of $Na2CO3$ and molecular weight of Magnesium sulphate. Themass of $Na2CO3$ mixture was found to be $x = 2.527 g$

The percent composition by mass of $Na2CO3$ in the mixture is:

$mass% Na2CO3 × mass Na2CO3mass of mixture × 100% = 33.1% Na2CO3$

The percent composition by mass of $Na2CO3$ in the mixture is calculated using the ratio between $mass Na2CO3$ and mass of mixture. The mass of $Na2CO3$ in the mixture is $33.1% Na2CO3$