Răspuns:
0,361 L etena si 0,01 L reactiv Bayer
Explicație:
[tex]3CH_{2}=CH_{2} + 2 KMnO_{4} + 4H_{2}O --> 3CH_{2}OH-CH_{2}OH + 2KOH + 2MnO_{2}\\M_{etan-1,2diol} = 2 * 12 + 6 * 1 + 16 * 2 = 24 + 6 + 32 = 62 \frac{g}{mol}[/tex]
3 moli etena .... 3 * 62 g produs
x moli etena ... 1 g
x = 0.016129 moli etena
[tex]PV=nRT\\V = \frac{nRT}{P}\\V = \frac{0.016129*0,082*273}{1}=0.361L[/tex]
2 moli Bayer .... 3 * 62 g produs
y moli Bayer .... 1 g produs
y = 0,01 moli Bayer
[tex]C = \frac{n}{V}\\V = \frac{n}{C}\\V = \frac{0,01}{1}=0,01L[/tex]
