Đáp án:

$a,$ $PTHH:$

$CaCO_3+2HCl→CaCl_2+CO_2+H_2O$ 

$2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$

$b,$ $V=0,672(l)$

$c,$ $x=1,88\ ;\ V'=2,24(l)$

$d,$

$C\%_{CaCl_2}=3,27\%$

$C\%_{HCl\ dư}=5,02\%$

Giải thích các bước giải:

$m_{HCl}=\dfrac{100×7,3}{100}=7,3(g)$

$n_{HCl}=\dfrac{7,3}{36,5}=0,2(mol)$

$m_{H_2SO_4}=\dfrac{100×9,8}{100}=9,8(g)$

$n_{H_2SO_4}=\dfrac{9,8}{98}=0,1(mol)$

$n_{CaCO_3}=\dfrac{3}{100}=0,03(mol)$

$n_{Al}=\dfrac{x}{27}(mol)$

$a,$

 $CaCO_3+2HCl→CaCl_2+CO_2+H_2O$ $(1)$

 $2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$ $(2)$

$b,$

Vì $\dfrac{n_{CaCO_3}}{1}<\dfrac{n_{HCl}}{2}⇒\ HCl$ dư

Theo PT $(1)$: $n_{CO_2}=n_{CaCO_3}=0,03(mol)$

$→V=0,03×22,4=0,672(l)$

$c,$

$TH1:\ H_2SO_4$ dư

Theo PT $(2)$: $n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{x}{18}$

Theo bài ra ta có: 

  $m_{Al}-m_{H_2}=m_{CaCO_3}-m_{CO_2}$

$⇒x-\dfrac{x}{18}×2=3-0,03×44$

$⇒x=1,89(g)$

$n_{Al}=\dfrac{1,89}{27}=0,07(mol)$

Xét $\dfrac{n_{Al}}{2}>\dfrac{n_{H_2SO_4}}{3}⇒\ TH1$ sai

$TH2:\ Al$ dư

Theo PT $(2)$: $n_{H_2}=n_{H_2SO_4}=0,1(mol)$

Theo bài ra ta có:

  $m_{Al}-m_{H_2}=m_{CaCO_3}-m_{CO_2}$

$⇒x-0,1×2=3-0,03×44$

$⇒x=1,88(g)$

$n_{Al}=\dfrac{1,88}{27}≈0,0696(mol)$
Xét $\dfrac{n_{Al}}{2}>\dfrac{n_{H_2SO_4}}{3}⇒\ TH2$ đúng

$→V'=0,1×22,4=2,24(l)$

$d,$

$m_{dd\ cốc\ 1}=3+100-0,03×44=101,68(g)$

Theo PT $(1)$: $n_{CaCl_2}=n_{CaCO_3}=0,03(mol)$

$n_{HCl\ (pư)}=2n_{CaCO_3}=2×0,03=0,06(mol)$

$n_{HCl\ (dư)}=0,2-0,06=0,14(mol)$

$→C\%_{CaCl_2}=\dfrac{0,03×111}{101,68}×100\%=3,27\%$

$→C\%_{HCl\ (dư)}=\dfrac{0,14×36,5}{101,68}×100\%=5,02\%$