Yechish kerak


Download 72 Kb.
bet2/4
Sana11.09.2023
Hajmi72 Kb.
#1675620
1   2   3   4
Bog'liq
10 ta masala yechimi

Yechish: Hosil bo`lgan eritma massasi: 1,4832 · 200 = 296,64
Sarflangan KOH eritmasi miqdori normal yoki 1,6 mol massasi

Demak eritmaga: 296,64 – 224 = 72,64 g oleum qo`shilgan ekan.
98x + 80y = 72,64 I – tenglama.
2x
xH2SO4 + 2KOH → K2SO4 + 2H2O
2y
ySO3 + 2KOH → K2SO4 + H2O
2x + 2y = 1,6 II – tenglama.
I va II tenglamalarni birgalikda yechamiz.
2x + 2y = 1,6 49
98x + 80y = 72,64
18y = 5,76
y = 0,32 mol SO3 uning massasi esa:
0,32 · 80 = 25,6
uning oleumdagi ω (%) : 25,6/72,64 · % = 35,25 %
Javob: 35,25 %.
A) 49.5 B) 50.5 C) 35.25 D) 64.75
3. Kalsiy oksid va kaliy oksiddan iborat 0.5 mol aralashma tarkibida kislorodga metallarning (yig’indisi) massa nisbati 1 : 2.975 ga teng bo’lsa, aralashmadagi kaliyning massa ulushini (%) aniqlang.
Yechish: Co Oni X, K2O ni y bilan belgilab tenglamalar sistemasi tuzamiz.
x CaO yK2O
tegishli ravishda metallar massasi 40x va 78y teng bo`ladi. Ularni yig`indisini 2,975 deb xisoblasak:
40x + 78y = 2,975 (I)
Kislorod massasi 1 ga teng bo`ladi:
16x + 16y = 1 (II)
I va II tenglamalarni birgalikda yechamiz:
40x + 78y = 2,975
16x + 16y = 1 2,5
40x + 40y = 2,5
40x + 78y = 2,975
38y = 0,475
y = 0,0125
kaliy massasi: 78y.
0,0125 · 0,975 g
aralashma massasi:
1 + 2,975 = 3,975
undagi kaliyning ω (%):
0,975/3,975 · 100% = 24,53%
Javob: 24,53 %.
A) 24.53 B) 33.6 C) 13.04 D) 49.5
4. Fe3O4,Fe2O3 va FeO aralashmasining og’irligi 69.6 g . Aralashmadagi kislorodlarning o’rtacha foizi 27.59 % bo’lsa, aralashmani qaytarish uchun kerak bo’ladigan vodorod hajmini (l.n.sh) toping.

Download 72 Kb.

Do'stlaringiz bilan baham:
1   2   3   4




Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling