Question

Volume 42litres of dinitrogen twmperture is 14 degree and preesure 1 atm find mass of sodium azide

Answer 1

Answer:

127 g NaN3

Explanation:

We're asked to find the mass, in g, of sodium azide (NaN3) needed to produce a certain amount of N2.

To do this, we can use the ideal gas equation to find the moles of nitrogen gas present:

PV=nRT

   P=1.15 atm (given)

   V must be in liters, so we can find the volume first in cm3 and then convert:

   50.0×50.0×25.0=62500 cm3

   62500cm3(1mL1cm3)(1lL103mL)=62.5 L

   R is the universal gas constant, equal to 0.082057L⋅atmmol⋅K

   T=25.0 oC, which must be in Kelvin:

   T=25.0 oC+273=298 K

Plugging in known values, and solving for the number of moles, n, we have

n=PVRT=(1.15atm)(62.5L)(0.082057L⋅atmmol⋅K)(298K)=2.94 mol N2

Now, we can use the coefficients of the chemical equation to find the relative number of moles of sodium azide that must react:

2.94mol N2(2lmol NaN33mol N2)=1.96 mol NaN3

Finally, we can use the molar mass of sodium azide (65.01 g/mol) to find the mass in grams:

1.96mol NaN3(65.01lg NaN31mol NaN3)=127lg NaN3−−−−−−−−−−

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