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Question: How many grams of Na(l) are produced per litre of \( {{{N}}_{{2}}} \) (g) formed in the decompositio...

How many grams of Na(l) are produced per litre of N2{{{N}}_{{2}}} (g) formed in the decomposition of sodium azide NaN3{{Na}}{{{N}}_{{3}}} if the gas is collected at 25 degree Celsius and 1.0{{1}}{{.0}} bar ( .98692{{.98692}} atm)? The R constant is 0.08206{{0}}{{.08206}} Latm/Kmol

Explanation

Solution

In the above question, we are asked to find out the weight of Na present per litre of N2{{{N}}_{{2}}} . So, here, we have to first write a balanced chemical equation of decomposition of sodium azide. Then we have to find the weight of Na present using the number of moles formula and volume of N2{{{N}}_{{2}}} occupied by the ideal gas equation. At the end, we have to find the ratio for grams per liter.

Formula Used
n=mM{{n = }}\dfrac{{{m}}}{{{M}}}
Where n = number of moles of gas
m = given mass
M = molar mass
PV=nRT{{PV = nRT}}
Where P = pressure.
V = volume
N = number of moles of gas
R = universal gas constant
T = temperature.

Complete step by step solution
Let us write a balanced equation of decomposition of sodium azide:
2NaN3→2Na+3N2{{2Na}}{{{N}}_{{3}}} \to {{2Na + 3}}{{{N}}_{{2}}}
Since, the number of moles of NaN3{{Na}}{{{N}}_{{3}}} , let us assume that it is equal to 1 mole.
From the above equation, we can infer that 2 moles of NaN3{{Na}}{{{N}}_{{3}}} form 2 moles of sodium.
So, 1 mole of NaN3{{Na}}{{{N}}_{{3}}} form 1 mole of sodium.
Weight of 1 mole of sodium = 23g.
Now, let us find out the volume of N2{{{N}}_{{2}}} .
Since, 2 moles of NaN3{{Na}}{{{N}}_{{3}}} form 3 moles of N2{{{N}}_{{2}}} .
So, 1 mole of NaN3{{Na}}{{{N}}_{{3}}} form 1.5{{1}}{{.5}} moles of N2{{{N}}_{{2}}} .
P = 1bar = 0.98692{{0}}{{.98692}} atm
R= 0.08206{{0}}{{.08206}} L atm/Kmol
T = 25+273=298{{25 + 273 = 298}} K
According to ideal gas equation, we have:
PV=nRT{{PV = nRT}}
Rearranging the above equation, we get:
V=nRTP{{V = }}\dfrac{{{{nRT}}}}{{{P}}}
Substituting the values, we have:
V=nRTP=1.5×0.08206×2980.98692=37.16{{V = }}\dfrac{{{{nRT}}}}{{{P}}}{{ = }}\dfrac{{{{1}}{{.5 \times 0}}{{.08206 \times 298}}}}{{{{0}}{{.98692}}}}{{ = 37}}{{.16}} L
Ratio:
Mass of Na : Volume of N2{{{N}}_{{2}}} = 2337.16=0.619\dfrac{{{{23}}}}{{{{37}}{{.16}}}}{{ = 0}}{{.619}}
Therefore, 0.619{{0}}{{.619}} g of Na is present in per liter of N2{{{N}}_{{2}}} .

Note
Ideal gas law is not applicable for real gas. But for numerical purposes, we generally use this law.
An equation must be balanced in order to satisfy the law of conservation of mass which states that mass can neither be created nor be destroyed.