`a)`
`n_{CaCO_3}={10}/{100}=0,1(mol)`
`n_{HCl}=0,5.2=1(mol)`
`CaCO_3+2HCl->CaCl_2+CO_2+H_2O`
Do `0,1<1/2->HCl` dư
`n_{HCl(du)}=1-0,1.2=0,8(mol)`
`n_{CaCl_2}=n_{CO_2}=0,1(mol)`
`m_{dd\ spu}=10+500.1,2-0,1.44=605,6(g)`
`C\%_{CaCl_2}={0,1.111}/{605,6}.100≈1,83\%`
`C\%_{HCl(du)}={0,8.36,5}/{605,6}.100≈4,82\%`
`b)`
`CO_2+NaOH->NaHCO_3`
`n_{NaOH}=n_{CO_2}=0,1(mol)`
`C_{M\ NaOH}={0,1}/{0,1}=1M`