   Chapter 10, Problem 100IL

Chapter
Section
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:

Total amount of the mixture = 1.232 gVolume of N2=295mL=295×10-3LPressure of N2=713mmHg=713mmHg760mmHg=0.938atmTemperature of N2= 21oC = 273.15K+21=278.15K

First using the ideal gas equation the moles of N2 produced is calculated as follows,

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

nN2=PVRT =(0.938atm)(295×10-3L)(0.0821L.atmK.mol)(278.15K)=0.01132molN2

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 0.01132molN2 should be then produced by 0.01132mol of gas present in the given mixture.

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

Let x be the mass of Na2CO3 in the mixture, then (2

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