Question #169060

NH4NO2 (s) ----> N2 (g) + H2O (g)

When a sample is decomposed in a test tube, 656** **mL of wet N_{2}(g) is collected over water at 28°C and 447 torr total pressure.

How many grams of dry NH_{4}NO_{2}(s) were initially decomposed? The vapor pressure of water at 28°C is 26.3 torr.

Mass? in g

Expert's answer

V= 656ml= 0.656L

T= 28°C= 28+273= 301K

P= 447torr

Recall that

760torr= 1atm

447torr= 1/760 x 447= 0.59atm

Vapour pressure of water= 26.3torr= 0.035atm

Partial pressure of N2= 0.59-0.035= 0.555atm

R= 0.0821L.atm/K.mol

Now let's find n, the number of moles of N2 from ideal gas equation.

PV=nRT

n= PV/RT

n= 0.555 x 0.656/ 0.0831 x 301

= 0.015mol

From the stoichiometry of reaction

1mol of N2 is produced from 1mol of NH4NO2

0.015mol of N2 will be produced from 0.015mol of NH4NO2

Mole of NH4NO2= 0.015mol

Molar mass of NH4NO2= 64g/mol

Mass of NH4NO2= mole x molar mass= 0.015 x 64= 0.96g of NH4NO2

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