Respuesta :
Answer : The mass of [tex]NaN_3[/tex] at temperature [tex]20^oC[/tex] 28.47 g.
The mass of [tex]NaN_3[/tex] at temperature [tex]10^oC[/tex] 29.51 g.
Solution : Given,
Pressure of gas = 1.35 atm
Temperature of gas = [tex]20^oC=273+20=293K[/tex] [tex](0^oC=273K)[/tex]
Volume of gas = [tex]25\times 25\times 20cm=12500cm^3=12.5L[/tex] [tex](1L=1000cm^3)[/tex]
Molar mass of [tex]NaN_3[/tex] = 65 g/mole
Part 1 : First we have to calculate the moles of gas at temperature [tex]20^oC[/tex]. The gas produced in the given reaction is [tex]N_2[/tex].
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
P = pressure of the gas
V = volume of the gas
T = temperature of the gas
n = number of moles of gas
R = Gas constant = 0.0821 Latm/moleK
Now put all the given values in this formula, we get
[tex](1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)[/tex]
By rearranging the terms, we get the value of 'n'
[tex]n=0.7015moles[/tex]
The moles of [tex]N_2[/tex] = 0.7015 moles
The given balanced reaction is,
[tex]20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)[/tex]
As, 32 moles of [tex]N_2[/tex] produced from 20 moles of [tex]NaN_3[/tex]
So, 0.7015 moles of [tex]N_2[/tex] produced from [tex]\frac{20}{32}\times 0.7015=0.438[/tex] moles of [tex]NaN_3[/tex]
Now we have to calculate the mass of [tex]NaN_3[/tex].
[tex]\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3[/tex]
[tex]\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g[/tex]
Therefore, the mass of [tex]NaN_3[/tex] needed are 28.47 g.
Part 2 : We have to calculate the moles of gas at temperature [tex]10^oC[/tex] and same volume & pressure.
Using ideal gas equation,
[tex]PV=nRT[/tex]
Now put all the given values in this formula, we get
[tex](1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)[/tex]
By rearranging the terms, we get the value of 'n'
[tex]n=0.726moles[/tex]
The moles of [tex]N_2[/tex] = 0.726 moles
The given balanced reaction is,
[tex]20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)[/tex]
As, 32 moles of [tex]N_2[/tex] produced from 20 moles of [tex]NaN_3[/tex]
So, 0.726 moles of [tex]N_2[/tex] produced from [tex]\frac{20}{32}\times 0.726=0.454[/tex] moles of [tex]NaN_3[/tex]
Now we have to calculate the mass of [tex]NaN_3[/tex].
[tex]\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3[/tex]
[tex]\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g[/tex]
Therefore, the mass of [tex]NaN_3[/tex] needed are 29.51 g.
