Răspuns :
1) n = [tex] \frac{68}{17} =[/tex] = 4 kmoli [tex]NH_{3} [/tex]
[tex]N_{2} + 3H_{2} = 2NH_{3}[/tex]
conform ecuatiei => 2 kmoli [tex]N_{2} [/tex] si 6 kmoli [tex]H_{2} [/tex] => 8 kmoli de gaze
pV = nRT => V = [tex] \frac{nRT}{p} = \frac{8*0,082*390}{4} = 63,96 m^{3} [/tex]
pentru [tex] NH_{3} [/tex] : V = [tex] \frac{4*0,082*390}{4} = 31,98m^{3} [/tex]
2) a) [tex]n_{1} = \frac{p_{1}V}{RT} = \frac{5*60}{0,082*293} [/tex] = 12,5 moli [tex]O_{2} [/tex]
[tex]n_{2}= \frac{p_{2}V}{RT} = \frac{2*60}{0,082*293} [/tex] = 5 moli [tex]O_{2} [/tex]
12,5 - 5 = 7,5 moli [tex]O_{2} [/tex] consumat => m = 7,5 · 32 = 240g [tex]O_{2} [/tex]
b) [tex]C_{2}H_{2} + \frac{5}{2} O_{2} = 2CO_{2} + H_{2}O [/tex]
1 mol acetilena ............. 2,5 moli oxigen
x moli acetilena ............. 7,5 moli oxigen
x = 3 moli acetilena
c) 5 moli oxigen ramas => V (c.n.) = 5 · 22,4 = 112 L oxigen
y · 60 = 112 => y = 1,866 ori
[tex]N_{2} + 3H_{2} = 2NH_{3}[/tex]
conform ecuatiei => 2 kmoli [tex]N_{2} [/tex] si 6 kmoli [tex]H_{2} [/tex] => 8 kmoli de gaze
pV = nRT => V = [tex] \frac{nRT}{p} = \frac{8*0,082*390}{4} = 63,96 m^{3} [/tex]
pentru [tex] NH_{3} [/tex] : V = [tex] \frac{4*0,082*390}{4} = 31,98m^{3} [/tex]
2) a) [tex]n_{1} = \frac{p_{1}V}{RT} = \frac{5*60}{0,082*293} [/tex] = 12,5 moli [tex]O_{2} [/tex]
[tex]n_{2}= \frac{p_{2}V}{RT} = \frac{2*60}{0,082*293} [/tex] = 5 moli [tex]O_{2} [/tex]
12,5 - 5 = 7,5 moli [tex]O_{2} [/tex] consumat => m = 7,5 · 32 = 240g [tex]O_{2} [/tex]
b) [tex]C_{2}H_{2} + \frac{5}{2} O_{2} = 2CO_{2} + H_{2}O [/tex]
1 mol acetilena ............. 2,5 moli oxigen
x moli acetilena ............. 7,5 moli oxigen
x = 3 moli acetilena
c) 5 moli oxigen ramas => V (c.n.) = 5 · 22,4 = 112 L oxigen
y · 60 = 112 => y = 1,866 ori