Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
Nh\u01b0 \u0111\u00e3 bi\u1ebft, ph\u00e2n t\u1eeb v\u00e0 nguy\u00ean t\u1eed c\u0169ng c\u00f3 k\u00edch th\u01b0\u1edbc v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng nh\u01b0ng c\u1ef1c k\u1ef3 nh\u1ecf b\u00e9, ch\u00fang ta kh\u00f4ng th\u1ec3 d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p th\u00f4ng th\u01b0\u1eddng \u0111\u1ec3 c\u00e2n \u0111o \u0111ong \u0111\u1ebfm \u0111\u01b0\u1ee3c. Nh\u01b0ng \u0111\u1ed1i v\u1edbi m\u00f4n h\u00f3a h\u1ecdc, ch\u00fang l\u1ea1i r\u1ea5t c\u1ea7n \u0111\u1ec3 t\u00ednh to\u00e1n ch\u00ednh x\u00e1c r\u1eb1ng c\u00f3 bao nhi\u00eau ph\u00e2n t\u1eed, nguy\u00ean t\u1eed v\u00e0 th\u1ec3 t\u00edch hay kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 bao nhi\u00eau. B\u00e0i vi\u1ebft sau s\u1ebd gi\u1ea3i \u0111\u00e1p cho b\u1ea1n kh\u00e1i ni\u1ec7m v\u1ec1 mol h\u00f3a 8 v\u00e0 c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp mol l\u1edbp 8.<\/p>\n\n\n\n
Mol l\u00e0 g\u00ec?<\/p>\n\n\n\n
Kh\u00e1i ni\u1ec7m mol trong mol h\u00f3a 8 nh\u01b0 sau:<\/p>\n\n\n\n
Mol l\u00e0 l\u01b0\u1ee3ng ch\u1ea5t ch\u1ee9a 6.1023 ph\u00e2n t\u1eed ho\u1eb7c nguy\u00ean t\u1eed c\u1ee7a ch\u1ea5t \u0111\u00f3<\/p>\n\n\n\n
Trong \u0111\u00f3: con s\u1ed1 6.1023 l\u00e0 s\u1ed1 Avogadro v\u00e0 \u0111\u01b0\u1ee3c k\u00fd hi\u1ec7u l\u00e0 N.<\/p>\n\n\n\n
Mol l\u00e0 g\u00ec?<\/p>\n\n\n\n
V\u00ed d\u1ee5:<\/p>\n\n\n\n
1 mol nguy\u00ean t\u1eed Fe = N nguy\u00ean t\u1eed Fe<\/p>\n\n\n\n
1 mol nguy\u00ean t\u1eed K = N nguy\u00ean t\u1eed K<\/p>\n\n\n\n
1 mol ph\u00e2n t\u1eed O2 = N ph\u00e2n t\u1eed O2<\/p>\n\n\n\n
1 mol ph\u00e2n t\u1eed N2 = N ph\u00e2n t\u1eed N2<\/p>\n\n\n\n
2 mol nguy\u00ean t\u1eed Cu = 2N nguy\u00ean t\u1eed Cu<\/p>\n\n\n\n
2 mol ph\u00e2n t\u1eed H2 = 2N ph\u00e2n t\u1eed H2<\/p>\n\n\n\n
3 mol nguy\u00ean t\u1eed Zn = 3N nguy\u00ean t\u1eed Zn<\/p>\n\n\n\n
4 mol ph\u00e2n t\u1eed O3 = 4N ph\u00e2n t\u1eed O3<\/p>\n\n\n\n
\u0110\u0103ng ky\u0301 ho\u0323c th\u01b0\u0309 mi\u00ea\u0303n phi\u0301 m\u00f4n Ho\u0301a ho\u0323c 8<\/p><\/blockquote>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng Mol l\u00e0 g\u00ec?<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol trong ch\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u1edbp 8 mol \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a nh\u01b0 sau:<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol (\u0111\u01b0\u1ee3c k\u00fd hi\u1ec7u l\u00e0 M) c\u1ee7a m\u1ed9t ch\u1ea5t n\u00e0o \u0111\u00f3 l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a N ph\u00e2n t\u1eed ho\u1eb7c nguy\u00ean t\u1eed ch\u1ea5t \u0111\u00f3. Kh\u1ed1i l\u01b0\u1ee3ng Mol \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng gam (g).<\/p>\n\n\n\n
K\u00fd hi\u1ec7u kh\u1ed1i l\u01b0\u1ee3ng mol: M<\/p>\n\n\n\n
\u0110\u01a1n v\u1ecb t\u00ednh: g\/mol<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol l\u00e0 g\u00ec?<\/p>\n\n\n\n
T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng mol \u0111\u01a1n ch\u1ea5t<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng ph\u00e2n t\u1eed hay nguy\u00ean t\u1eed c\u1ee7a m\u1ed9t \u0111\u01a1n ch\u1ea5t n\u00e0o \u0111\u00f3 c\u00f3 c\u00f9ng tr\u1ecb s\u1ed1 v\u1edbi ph\u00e2n t\u1eed kh\u1ed1i hay nguy\u00ean t\u1eed kh\u1ed1i c\u1ee7a ch\u1ea5t \u0111\u00f3.<\/p>\n\n\n\n
V\u00ed d\u1ee5 v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng mol \u0111\u01a1n ch\u1ea5t:<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a nguy\u00ean t\u1eed \u0110\u1ed3ng l\u00e0 MCu = 64 g\/mol<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a nguy\u00ean t\u1eed S\u1eaft l\u00e0 MFe = 56 g\/mol<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a ph\u00e2n t\u1eed Nit\u01a1 l\u00e0 MN2 = 28 g\/mol<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a ph\u00e2n t\u1eed Kali l\u00e0 MK2 = 78 g\/mol<\/p>\n\n\n\n
T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng mol ph\u00e2n t\u1eed, h\u1ee3p ch\u1ea5t<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a m\u1ed9t h\u1ee3p ch\u1ea5t, ph\u00e2n t\u1eed bao g\u1ed3m nguy\u00ean t\u1eed c\u1ee7a c\u00e1c nguy\u00ean t\u1ed1 kh\u00e1c nhau. \u0110\u01b0\u1ee3c t\u00ednh b\u1eb1ng t\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a nguy\u00ean t\u1eed c\u00f3 trong h\u1ee3p ch\u1ea5t, ph\u00e2n t\u1eed ch\u1ea5t \u0111\u00f3.<\/p>\n\n\n\n
Tr\u01b0\u1eddng h\u1ee3p t\u1ed5ng qu\u00e1t: V\u1edbi b\u1ea5t k\u1ef3 m\u1ed9t h\u1ee3p ch\u1ea5t n\u00e0o \u0111\u00f3 \u0111\u01b0\u1ee3c c\u1ea5u t\u1ea1o b\u1edfi 2 nguy\u00ean t\u1ed1 h\u00f3a h\u1ecdc n\u00e0o \u0111\u00f3 v\u1edbi c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc k\u00fd hi\u1ec7u l\u00e0 AaBb trong \u0111\u00f3:<\/p>\n\n\n\n
A, B l\u00e0 nguy\u00ean t\u1ed1 c\u1ea5u t\u1ea1o n\u00ean h\u1ee3p ch\u1ea5t AaBb<\/p>\n\n\n\n
a, b l\u00e0 s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a nguy\u00ean t\u1ed1 t\u01b0\u01a1ng \u1ee9ng<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a h\u1ee3p ch\u1ea5t AaBb \u0111\u01b0\u1ee3c t\u00ednh nh\u01b0 sau:<\/p>\n\n\n\n
Mhc = MA x a + MB x b<\/p>\n\n\n\n
V\u00ed d\u1ee5: H\u00e3y t\u00ednh kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a c\u00e1c h\u1ee3p ch\u1ea5t sau: mu\u1ed1i \u0103n: NaCl, kh\u00ed Metan: CH4.<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a mu\u1ed1i \u0103n (NaCl) l\u00e0:<\/p>\n\n\n\n
MNaCl = MNa + MCl = 23 + 35,5 = 58,5 g\/mol<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a kh\u00ed metan CH4 l\u00e0:<\/p>\n\n\n\n
MCH4 = MC + MH4 = 12 + (1\u00d74) = 16 g\/mol<\/p>\n\n\n\n
L\u01b0u \u00fd:<\/p>\n\n\n\n
\u2013 Kh\u1ed1i l\u01b0\u1ee3ng mol th\u01b0\u1eddng xuy\u00ean g\u1eb7p trong c\u00e1c b\u00e0i t\u1eadp t\u00ednh to\u00e1n<\/p>\n\n\n\n
\u2013 N\u00ean quy \u0111\u1ed5i kh\u1ed1i l\u01b0\u1ee3ng t\u00ednh ra mol \u0111\u1ec3 t\u00ednh to\u00e1n<\/p>\n\n\n\n
Th\u1ec3 t\u00edch mol ch\u1ea5t kh\u00ed l\u00e0 g\u00ec?<\/p>\n\n\n\n
Kh\u00e1i ni\u1ec7m v\u1ec1 th\u1ec3 t\u00edch mol ch\u1ea5t kh\u00ed trong mol h\u00f3a 8 nh\u01b0 sau:<\/p>\n\n\n\n
Th\u1ec3 t\u00edch mol ch\u1ea5t kh\u00ed l\u00e0 th\u1ec3 t\u00edch chi\u1ebfm N ph\u00e2n t\u1eed ch\u1ea5t kh\u00ed \u0111\u00f3. M\u1ed9t mol ch\u1ea5t kh\u00ed c\u1ee7a b\u1ea5t k\u1ef3 ch\u1ea5t n\u00e0o, trong c\u00f9ng m\u1ed9t \u0111i\u1ec1u ki\u1ec7n v\u1ec1 \u00e1p su\u1ea5t v\u00e0 nhi\u1ec7t \u0111\u1ed9 \u0111\u1ec1u chi\u1ebfm l\u01b0\u1ee3ng th\u1ec3 t\u00edch b\u1eb1ng nhau.<\/p>\n\n\n\n
Th\u1ec3 t\u00edch mol ch\u1ea5t kh\u00ed l\u00e0 g\u00ec?<\/p>\n\n\n\n
N\u1ebfu \u1edf \u0111i\u1ec1u ki\u1ec7n \u00e1p su\u1ea5t l\u00e0 1 atm v\u00e0 nhi\u1ec7t \u0111\u1ed9 l\u00e0 0 \u0111\u1ed9 C th\u00ec \u0111\u1ec1u c\u00f3 th\u1ec3 t\u00edch mol l\u00e0 22,4 l\u00edt. \u0110\u00e2y \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n v\u00e0 vi\u1ebft t\u1eaft l\u00e0 \u0111ktc. Nh\u01b0 v\u1eady, c\u00e1c ch\u1ea5t kh\u00ed kh\u00e1c nhau c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol kh\u00e1c nhau nh\u01b0ng th\u1ec3 t\u00edch mol c\u1ee7a ch\u00fang l\u1ea1i gi\u1ed1ng nhau khi c\u00f3 c\u00f9ng \u0111i\u1ec1u ki\u1ec7n v\u1ec1 \u00e1p su\u1ea5t v\u00e0 nhi\u1ec7t \u0111\u1ed9.<\/p>\n\n\n\n
V\u00ed d\u1ee5:<\/p>\n\n\n\n
\u2013 \u1ede \u0111ktc, 1 mol kh\u00ed metan CH4 ho\u1eb7c 1 mol kh\u00ed hidro H2 \u0111\u1ec1u c\u00f3 th\u1ec3 t\u00edch nh\u01b0 nhau l\u00e0 22,4 l\u00edt.<\/p>\n\n\n\n
\u2013 \u1ede \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng, \u00e1p su\u1ea5t 1 atm v\u00e0 nhi\u1ec7t \u0111\u1ed9 20 \u0111\u1ed9 C th\u00ec 1 mol ch\u1ea5t kh\u00ed b\u1ea5t k\u1ef3 c\u00f3 th\u1ec3 t\u00edch l\u00e0 22 l\u00edt.<\/p>\n\n\n\n
\u0110\u00e2y l\u00e0 2 \u0111i\u1ec1u ki\u1ec7n quan tr\u1ecdng \u0111\u1ec3 l\u00e0m b\u00e0i t\u1eadp t\u00ednh to\u00e1n, c\u00e1c b\u1ea1n n\u00ean l\u01b0u \u00fd k\u1ef9 nh\u00e9.<\/p>\n\n\n\n
Vietlearn n\u1ec1n t\u1ea3ng h\u1ecdc tr\u1ef1c tuy\u1ec3n \u00e1p d\u1ee5ng c\u00f4ng ngh\u1ec7 AI nh\u1eafc h\u1ecdc th\u00f4ng minh<\/p><\/blockquote>\n\n\n\n
T\u1ed5ng h\u1ee3p c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u00e0 l\u1eddi gi\u1ea3i mol h\u00f3a 8<\/p>\n\n\n\n
C\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft c\u1ee7a mol l\u1edbp 8 \u0111\u01b0\u1ee3c t\u1ed5ng h\u1ee3p d\u01b0\u1edbi \u0111\u00e2y, m\u1eddi b\u1ea1n \u0111\u00f3n \u0111\u1ecdc.<\/p>\n\n\n\n
T\u1ed5ng h\u1ee3p c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u00e0 l\u1eddi gi\u1ea3i mol h\u00f3a 8<\/p>\n\n\n\n
D\u1ea1ng 1: T\u00ednh s\u1ed1 mol ph\u00e2n t\u1eed ho\u1eb7c s\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a nh\u1eefng l\u01b0\u1ee3ng ch\u1ea5t sau:<\/p>\n\n\n\n
a. 2,4.1023 nguy\u00ean t\u1eed Fe<\/p>\n\n\n\n
b. 24 gam NaOH<\/p>\n\n\n\n
c. 11,2 l\u00edt kh\u00ed H2 (\u0111ktc)<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/p>\n\n\n\n
a) S\u1ed1 mol nguy\u00ean t\u1eed trong 2.1023 nguy\u00ean t\u1eed Fe l\u00e0:<\/p>\n\n\n\n
b) S\u1ed1 mol NaOH c\u00f3 trong 24 gam NaOH l\u00e0:<\/p>\n\n\n\n
c) S\u1ed1 mol ph\u00e2n t\u1eed c\u00f3 trong 11,2 l\u00edt kh\u00ed H2 (\u0111ktc) l\u00e0:<\/p>\n\n\n\n
D\u1ea1ng 2: T\u00ednh s\u1ed1 mol ch\u1ea5t tan trong 400 ml dung d\u1ecbch NaCl 0,6M<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/p>\n\n\n\n
\u0110\u1ed5i 400 ml = 0,4 l\u00edt<\/p>\n\n\n\n
S\u1ed1 mol ch\u1ea5t tan c\u00f3 trong 400 ml dung d\u1ecbch NaCl 0,6M l\u00e0:<\/p>\n\n\n\n
n = CM x V = 0,6 x 0,4 = 0,24 mol<\/p>\n\n\n\n
D\u1ea1ng 3: T\u00ednh s\u1ed1 mol FeCl2 c\u00f3 trong 400 gam dung d\u1ecbch FeCl2 19,05%<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/p>\n\n\n\n
Kh\u1ed1i l\u01b0\u1ee3ng FeCl2 trong dung d\u1ecbch l\u00e0:<\/p>\n\n\n\n
S\u1ed1 mol FeCl2 trong dung d\u1ecbch l\u00e0:<\/p>\n\n\n\n
B\u00e0i vi\u1ebft tr\u00ean \u0111\u00e2y \u0111\u00e3 cung c\u1ea5p cho b\u1ea1n \u0111\u1ecdc nh\u1eefng kh\u00e1i ni\u1ec7m v\u1ec1 mol trong mol h\u00f3a 8 v\u00e0 c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp t\u1ed5ng h\u1ee3p. Hy v\u1ecdng b\u00e0i vi\u1ebft s\u1ebd gi\u00fap \u00edch cho b\u1ea1n trong qu\u00e1 tr\u00ecnh h\u1ecdc t\u1eadp<\/a> v\u00e0 nghi\u00ean c\u1ee9u. Ch\u00fac b\u1ea1n th\u00e0nh c\u00f4ng!<\/p>\n\n\n\n
<\/p>\n\n\n\n
Chuy\u1ec3n \u0111\u1ed5i gi\u1eefa kh\u1ed1i l\u01b0\u1ee3ng, th\u1ec3 t\u00edch v\u00e0 ch\u1ea5t l\u01b0\u1ee3ng H\u00f3a 8<\/p>\n\n\n\n
T\u1ec9 kh\u1ed1i c\u1ee7a ch\u1ea5t kh\u00ed: c\u00f4ng th\u1ee9c v\u00e0 c\u00e1c d\u1ea1ng b\u00e0<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":"
Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn Nh\u01b0 \u0111\u00e3 bi\u1ebft, ph\u00e2n t\u1eeb v\u00e0 nguy\u00ean t\u1eed c\u0169ng c\u00f3 k\u00edch th\u01b0\u1edbc v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng nh\u01b0ng c\u1ef1c k\u1ef3 nh\u1ecf b\u00e9, ch\u00fang ta kh\u00f4ng th\u1ec3 d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p th\u00f4ng th\u01b0\u1eddng \u0111\u1ec3 c\u00e2n \u0111o \u0111ong \u0111\u1ebfm \u0111\u01b0\u1ee3c. Nh\u01b0ng \u0111\u1ed1i v\u1edbi m\u00f4n h\u00f3a h\u1ecdc, ch\u00fang l\u1ea1i r\u1ea5t c\u1ea7n \u0111\u1ec3 t\u00ednh to\u00e1n ch\u00ednh […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_disable_autopaging":false},"categories":[8,4],"tags":[],"class_list":["post-770","post","type-post","status-publish","format-standard","hentry","category-hoa-hoc","category-trung-hoc"],"_links":{"self":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/770","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/comments?post=770"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/770\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=770"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=770"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=770"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}