Đáp án + Giải thích các bước giải:

 a. $m_{Fe} = 160 - 3 . 16 = 160 - 48 = 112g$

$x = n_{Fe} = \dfrac{m}{M} = \dfrac{112}{56} = 2(\text{mol})$

$\Rightarrow CTHH: Fe_2O_3$

b. $m_{SO_4} = 342 - 2 . 27 = 342 - 54 = 288g$

$x = n_{SO_4} = \dfrac{m}{M} = \dfrac{288}{32  + 16 . 4} = 3(\text{mol})$

$\Rightarrow CTHH: Al_2(SO_4)_3$

c. $m_{NO_3} = 188 - 64 = 124g$

$x = n_{NO_3} = \dfrac{m}{M} = \dfrac{124}{14 + 3 . 16} = 2(\text{mol})$

$\Rightarrow CTHH: Cu(NO_3)_2$

$Đáp$ $án$ $+$ $Giải$ $thích$ $các$ $bước$ $giải$ $:$

$a,$

$M_{Fe_x O_3}$ $=$ $160$

$⇒$ $56x$ $+$ $16×3$ $=$ $160$ 

$⇒$ $x=2$ 

$⇒$ $CTHH$ $:$ $Fe_2 O_3$

$b,$

$M_{Al_2 (SO_4)_x}$ $=$ $342$

$⇒$ $27×2$ $+$ $(32+16×4)x$ $=$ $342$ 

$⇒$ $x=3$ 

$⇒$ $CTHH$ $:$ $Al_2 (SO_4)_3$

$c,$

$M_{Cu(NO_3)_x}$ $=$ $188$

$⇒$ $64$ $+$ $(14+16×3)x$ $=$ $188$ 

$⇒$ $x=2$ 

$⇒$ $CTHH$ $:$ $Cu(NO_3)_2$

$Chúc$ $bn$ $hk$ $tốt$ $!!$ $:))))$

$Xin$ $ctlhn$ $ạ.$