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`a)`

`448ml=0,448l`

`n_{CO_2}={0,448}/{22,4}=0,02(mol)`

`n_{CaCl_2}={3,33}/{111}=0,03(mol)`

`CaCO_3+2HCl->CaCl_2+CO_2+H_2O`

`0,02←0,04←0,02←0,02(mol)`

`CaO+2HCl->CaCl_2+H_2O`

`0,01←0,02←0,03-0,02=0,01(mol)`

`\%m_{CaCO_3}={0,02.100}/{0,02.100+0,01.56}.100\%=78,125\%`

`\%m_{CaO}=100-78,125=21,875\%`

`b)`

`n_{HCl}=0,04+0,02=0,06(mol)`

`->V_{dd\ HCl}={0,06}/{0,2}=0,3(l)=300(ml)`

`->m_{dd\ HCl}=300.1,1=330(g)`

`->m_{dd\ spu}=0,02.100+0,01.56+330-0,02.44=331,68(g)`

`->C\%_{CaCl_2}={3,33}/{331,68}.100\%≈1\%`

`a)`

`CaO+ 2HCl→ CaCl_2+ H_2O` 

`CaCO_3+ 2HCl→ CaCl_2+ CO_2+ H_2O` 

`448ml=0,448l` 

`n_{CO_2}= {0,448}/{22,4}=0,02(mol)` 

`n_{CaCO_3}= n_{CO_2}= 0,02(mol)` 

`n_{CaCl_2}={3,33}/{111}=0,03(mol)` 

Bảo toàn nguyên tố Ca: 

`n_{CaCl_2}= n_{CaO}+ n_{CaCO_3}`

`→ n_{CaO}=0,01(mol)` 

`%m_{CaO}={0,01. 56}/{0,01. 56+0,02. 100}. 100=21,875%`

`%m_{CaCO_3}=100- 21,875=78,125%`

`b)` 

`n_{HCl}= 2n_{CaCl_2}=0,06(mol)` 

`V= {0,06}/{0,2}=0,3(l)=300(ml)` 

`mdd_{HCl}= 300. 1,1=330(g)`

`mdd=330+0,01. 56+0,02. 100-0,02. 44=331,68(g)`

`C%_{CaCl_2}={3,33}/{331,68}. 100=1%`