Gjest skrev:hvordan balansere
[tex]N_2H_4O_3+C_{16}H_{34}\leftrightarrow CO_2+H_2O+N_2[/tex]
?
Litt stress å sammenligne koeffisienter uten videre her;
men:
- [tex]\alpha N_2H_4O_3+\beta C_{16}H_{34}\Longrightarrow \gamma CO_2+\delta H_2O+\epsilon N_2[/tex]
[tex]2\alpha =2\epsilon \,\,\,\,\, \left ( I \right )[/tex]
[tex]4\alpha +34\beta =2 \delta \,\,\,\,\, (II)[/tex]
[tex]3\alpha =2\gamma +\delta \,\,\,\,\, (III)[/tex]
[tex]16\beta =\gamma \,\,\,\,\, (III)[/tex]
Fra [tex](I)[/tex] har vi at [tex]\alpha=\epsilon[/tex]. Setter [tex]\alpha =\epsilon =1[/tex]
Rydder litt opp i [tex](II)[/tex] og [tex](III)[/tex]
[tex]4\alpha +34\beta =2\delta \Longrightarrow 4+34\beta =2\delta \,\,\,\,\, (II)[/tex]
[tex]3\alpha =2\gamma +\delta \Longrightarrow 3=2\gamma +\delta \,\,\,\,\, (III)[/tex]
Tre likninger med tre ukjente
- [tex]\star_1: \,\,\,\,\, 2+17\beta =\delta[/tex]
- [tex]\star_2: \,\,\,\,\, 3=2\gamma +\delta[/tex]
- [tex]\star_3: \,\,\,\,\,\,\,\,\,\, 16\beta =\gamma[/tex]
Det gir:
[tex]L:\left \{ \beta =\frac{1}{49},\delta =\frac{115}{49},\gamma =\frac{16}{49} \right \}[/tex]
MAO:
[tex]\Large N_2H_4O_3+\frac{1}{49}C_{16}H_{34}\Longrightarrow \frac{16}{49}CO_2+\frac{115}{49}H_2O+N_2[/tex]
Ved multiplisering med felles nevner [tex]49[/tex] får vi at:
[tex]\Large 49N_2H_4O_3+C_{16}H_{34}\Longrightarrow 16CO_2+115H_2O+49N_2[/tex]