Chapter 10, Problem 100IL

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# You are given a solid mixture of NaNO2 and NaCl and are asked to analyze it for the amount of NaNO2 present. To do so, you allow the mixture to react with sulfamic acid, HSO3NH2, in water according to the equationNaNO2(aq) + HSO3NH2(aq) →NaHSO4(aq) + H2O(ℓ) + N2(g)What is the weight percentage of NaNO2 in 1.232 g of the solid mixture if reaction with sulfa-mic acid produces 295 mL of dry N2 gas with a pressure of 713 mm Hg at 21.0 °C?

Interpretation Introduction

Interpretation:

The percent composition by mass of NaNO2 in the mixture should be calculated.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas.  Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained.  It is referred as ideal gas equation.

nTPV = RnTPPV = nRTwhere,n = moles of gasP = pressureT = temperatureR = gas constant

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties.  At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Explanation

Given:

â€‚Â TotalÂ amountÂ ofÂ theÂ mixtureÂ =Â 1.232Â gVolumeÂ ofÂ N2â€Šâ€Šâ€Šâ€Šâ€Š=â€Šâ€Šâ€Š295mLâ€Šâ€Šâ€Šâ€Š=â€Šâ€Šâ€Šâ€Š295Ã—10-3LPressureÂ ofÂ N2â€Šâ€Šâ€Šâ€Š=â€Š713mmHgâ€Šâ€Š=â€Šâ€Š713mmHg760mmHgâ€Šâ€Š=â€Šâ€Š0.938atmTemperatureÂ ofÂ N2â€Šâ€Šâ€Šâ€Š=Â 21oCÂ =Â 273.15K+21â€Šâ€Š=â€Šâ€Š278.15K

First using the ideal gas equation the moles of N2â€Š produced is calculated as follows,

Here, we carry an extra significant figure throughout this calculation to limit rounding errors.

â€‚Â Â nN2â€‰=PVRTÂ â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰(â€Š0.938atm)(295Ã—10-3L)(0.0821Lâ€‰.â€‰atmKâ€‰.â€‰mol)â€‰(278.15K)â€‰â€‰=â€‰0.01132â€‰molâ€‰N2

From the balanced chemical equation it is clear that one mole of gas gives rise to one mole of N2 hence, the total calculated amount of 0.01132â€‰molâ€‰N2 should be then produced by 0.01132â€‰molâ€‰ of gas present in the given mixture.

Hence, there exists 1:1 mole ratio between N2 and NaNO2 which show that 0.01132â€‰molâ€‰ from the mixture must have reacted.

LetÂ xÂ beÂ theÂ massÂ ofÂ Na2CO3Â inÂ theÂ mixture,Â thenÂ (2

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