`#Ly`

Bài `bb2:`

$a)$

Đổi $200$ $ml$ $=$ $0,2$ $l$

`n_(HNO_3` `=` `0,5.0,2=0,1` $(mol)$

`m_(HNO_3` `=` `0,1.63=6,3` $(g)$

$b)$

`m_(Ca(OH)_2` `=` `(150.7,4)/(100)=11,1` $(g)$

`n_(Ca(OH)_2` `=` `(11,1)/(74)=0,15` $(mol)$

$c)$

`n_(Fe_2(SO_4)_3` `=` `0,2.4=0,8` $(mol)$

`m_(Fe_2(SO_4)_3` `=` $0,8.400=320` $(g)$

$d)$

Đổi $0,08$ $kg$ $=$ `80` `g`

`m_(Ba(OH)_2` `=` `(80.17,1)/(100)=13,68` $(g)$

`n_(Ba(OH)_2` `=` `(13,68)/(171)=0,08` $(mol)$

Bài `bb3:`

$a)$

`n_(Zn` `=` `(13)/(65)=0,2` $(mol)$

PTHH: $Zn$ $+$ $H_2SO_4$ `->` $ZnSO_4$ $+$ $H_2$

`->` `n_(H_2` `=` `n_(Zn` `=` `0,2` $(mol)$

`V_(H_2` `=` `0,2.22,4=4,48` $(l)$

$b)$

Theo PT, `n_(H_2SO_4` `=` `n_(Zn` `=` `0,2` $(mol)$

`m_(H_2SO_4` `=` `0,2.98=19,6` $(g)$

`m_(ddH_2SO_4` `=` `(19,6.100)/(19,6)=100` $(g)$

`m_(ddZnSO_4` `=` `13+100-0,2.2=112,6` $(g)$

Theo PT, `n_(ZnSO_4` `=` `n_(Zn` `=` `0,2` $(mol)$

`C%_(ZnSO_4` `=` `(0,2.161)/(112,6),100%≈28,6%`

Đáp án:

Giải thích các bước 

a)nHNO3=0.5.0.2=0.1 mol 

mHNO3=0.1.(1+14+16.3)=6.3(g)

b)nca(oh)=7,4 . 0,15 =1.11 mol 

mCaoh=1.11 . ( 40 + 16 + 1 ) =63.27(g)

c)nfe2(so4)3=0.2.4=0.8 (mol)

mfe2(so4)3=0.8 . ( 56.2 + 32.3 + 16.12)=192g)

d)nba(oh)=0.08 . 17.1 = 1.368(mol)

mba(oh)=1.368 . ( 137+ 17)=210.672(g)

///////////////////////////////////////////

a)nZn=13 : 65 = 0.2(mol)

Zn+h2so4-Znso4+h2

0,2     0.2    0.2       0.2 mol

mh2=0,2 . 2 = 0.4(g)

vh2=0.2.22.4=4.48(l)

b)mZnso4=0.2 . (65+32 +16.4)=32.2(g)

mh2so4=0.2 . (2+32+16.4)=19.6(g)

mdd=(19.6/19.6:100)+13 -0.4=112.6(g)

cm=32.2/112.6 . 100%=28.60%