The initial reaction in an automobile airbag is : 2NaN3(s) → 3N2(g) + 2Na(s)
How many liters of N2 at 25.0oC and 1.00 atm. result when 105grams of NaN3 react?
2007-04-27 00:56:31 · 3 answers · asked by JULIO C 1 in Science & Mathematics ➔ Chemistry
2NaN3(s) → 3N2(g) + 2Na(s)
From this 2 moles of sodium nitride produces 3 moles of nitrogen. The molar mass of sodium nitride is 65.00999g/mol and therefore the number of moles of sodium nitride we have is 105/65.00999= 1.6151moles. This will produce 1.6151*3/2= 2.4227 moles of Nitrogen.
Using the ideal gas law we can calculate the volume of nitrogen produced.
PV=nRT
V= nRT/P= 2.4227*8.314*(25+273)/(1.013 x10^5) = 0.05925m^3= 59.25litres
2007-04-27 02:24:20 · answer #1 · answered by The exclamation mark 6 · 0⤊ 0⤋
first of all calculate moles of NaN3
its molecular mass is: 14*3+23= 65 g/mol
n=g/MM
so 105/65= 1.61 mol
u know that 2 moles of NaN3 give 3 moles of nitrogen, so
2:3=1.61: x
x=2.41 mol
so just apply PV=nRT
where P is pressure, V is Volume (that is what u have to calculate), n are moles, R=0.0821 and T is temperature expressed in Kelvin...so just have to calculate it.... :)
2007-04-27 02:23:27 · answer #2 · answered by Lyla 6 · 0⤊ 0⤋
47.5 litres.
2007-04-27 21:39:43 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋