La 1) te rog verifica datele problemei daca le-ai scris corect.....!!!!!!!!!!!!!!
1).
n moli amestec = 36,132 * 10^22 molecule / 6,0225 * 10^23 molecule
= 0,6 moli
a moli n g b g
CO2 + 2NaOH => Na2CO3 + H2O
44 2x40 106
c moli m g d g
SO3 + 2NaOH => Na2SO4 + H2O
80 2x40 142
a + c = 0,6
b/d = 0,373/1 => b = 0,373d
b = 106a/44 = 2,41a
d = 142c/80 = 1,775c
=> 2,41a = 0,373x1,775c = 0,662c
=> 3,64a = c
=> a + 3,64a = 0,6 => a = 0,13 moli CO2
diferenta pana la 0,6 = 0,47 moli SO3
din reactii avem:
n = 2x40x0,13/44 = 0,236 g NaOH
m = 2x40x0,47/80 = 0,47 g NaOH
m(NaOH) = 0,236+0,47 = 0,706 g
stim ca c% = mdx100/ms
=> ms = mdx100/c% = 0,706x100/20 = 3,53 g sol. NaOH
ro = p = ms/Vs => Vs = ms/p
= 3,53/1,225 = 2,88 mL sol. NaOH
2).
A.
100% aer ....................... 20% O2
la 11,2 L ............................ V(o2) = 2,24 L
n moli 2,24 L m g
2Mg + O2 => 2MgO
2 22,4 2x40
=> m = 2,24x2x40/22,4 = 80 g MgO => Fals
B.
n = 2x2,24/22,4 = 0,2 moli Mg
1 mol Mg .................. 1x6,0225*10^23 atomi Mg
0,2 moli Mg.............. c = 1,2044*10^23 sau 0,12044*10^24
=> Adevarat
C.
din cei 11,2 L aer diferenta este 8,96 L N2
22,4 L N2 ................... 1*NA atomi N
8,96 L n2 ...................... d = 0,4*NA => Adevarat
