0Bình chọn giảmTrong mỗi phần có khối lượng là 12,12" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">12,1212,12= 6,05(g) 4Al+3O2→2Al2O3" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">4Al+3O2→2Al2O34Al+3O2→2Al2O3(mol) a : 0,75a 2Mg+O2→2MgO" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2Mg+O2→2MgO2Mg+O2→2MgO(mol) b : 0,5b 4Na+O2→2Na2O" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">4Na+O2→2Na2O4Na+O2→2Na2O(mol) c : 0,25cTheo định luật bảo toàn khối lượng: mKimloại+mO2=mOxit" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">mKimloại+mO2=mOxitmKimloại+mO2=mOxit⇒6,05+mO2=9,65" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">⇒6,05+mO2=9,65⇒6,05+mO2=9,65 ⇒mO2=3,6(g)⇒nO2=0,1125(mol)" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">⇒mO2=3,6(g)⇒nO2=0,1125(mol)⇒mO2=3,6(g)⇒nO2=0,1125(mol)⇒0,7a+0,5b+0,25c=0,1125" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">⇒0,7a+0,5b+0,25c=0,1125⇒0,7a+0,5b+0,25c=0,1125 2Al+6HCl→2AlCl3+3H2" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2(mol)a : 3a : :1,5a Mg+2HCl→MgCl2+H2" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">Mg+2HCl→MgCl2+H2Mg+2HCl→MgCl2+H2(mol)b : 2b : : b 2Na+2HCl→2NaCl+H2" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2Na+2HCl→2NaCl+H22Na+2HCl→2NaCl+H2(mol)c : c : : 0,5aTa có:nH2=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">nH2=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)nH2=1,5a+b+0,5c=2.(0,75a+0,5b+0,25c)=2.0,1125=0,225(mol)⇒V=0,225.22,4=5,04(l)" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">⇒V=0,225.22,4=5,04(l)⇒V=0,225.22,4=5,04(l)nCl=nHCl=2nH2=0,225.2=0,45(mol)" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">nCl=nHCl=2nH2=0,225.2=0,45(mol)nCl=nHCl=2nH2=0,225.2=0,45(mol)⇒m=mkl+mCl=6,05+0,45.35,5=22,025(g)" role="presentation" style="font-size: 12px; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">⇒m=mkl+mCl=6,05+0,45.35,5=22,025(g)@shiro
0Bình chọn giảm
Đáp
án
: Gọi
x,y
,z l
ần l
ượt l
à s
ố mol c
ủa
Al
, Mg
, Na( x
,y,z>0)4Al+3O2--
> 2Al2O3x-------------
2/4x (mol)2
Mg+O2--
> 2
MgOy--------------
y(mol)
4Na+
O2--
> 2
Na2
Oz--------------2
/4zm m
ỗi ph
ần =12,
1:2=
6,05theo
bài ra ta c
ó : 27
x+2
4y+2
3z=
6,05(
g)102.2
/4x
+40
y+62.2
/4z=
9,65(g)