* Đề 5:
1)
a, 3Fe+ 2O2 (t*)= Fe3O4
3:2:1
b, AlCl3+ 3NaOH= Al(OH)3+ 3NaCl
1:3:1:3
c, BaO+ 2HCl= BaCl2+ H2O
1:2:1:1
d, Fe2O3+ 6HNO3= 2Fe(NO3)3+ 3H2O
1:6:2:3
2)
Áp dụng QTHT, ta có:
a, Zn(OH)2
b, Al(NO3)3
c, MgCl2
3)
a, 1,8.10^23= 0,3 mol NaCl
=> mNaCl= 0,3.58,5= 17,55g
b, nCO2= 11,2/22,4= 0,5 mol
=> mCO2= 0,5.44= 22g
c, m= 0,2.32+ 0,4.28= 17,6g
4)
Na2CO3 có:
%Na= 23.2.100:106= 43,4%
%C= 12.100:106= 11,32%
%O= 100-43,4-11,32= 45,28%
5)
a, K2O+ H2O= 2KOH
mK2O+ mH2O= mKOH
b, mKOH= 9,4+1,8= 11,2g
* Đề 7:
1)
a, 2C4H10 + 13 O2 (t*)= 8 CO2+ 10H2O
2: 13: 8: 10
b, 2Fe(OH)3 (t*)= Fe2O3+ 3H2O
2:1:3
c, 2K+ 2H2O= 2KOH+ H2
2:2:2:1
d, Al2O3+ 6HCl= 2AlCl3+ 3H2O
1:6:2:3
2)
MgCl2, Na2O, Ca(OH)2, ZnSO4
3)
a, n= 42,8/107= 0,4 mol
b, n= 9.10^23 / 6.20^23= 1,5 mol
c, n= 8,4/28= 0,3 mol
=> V C2H4= 0,3.22,4= 6,72l
d, m= 0,3.102+ 0,15.81= 42,75g
