`a)`
`n_{MnO_2}={78,3}/{87}=0,9(mol)`
`MnO_2+4HCl->MnCl_2+Cl_2+2H_2O`
`n_{HCl}=4n_{MnO_2}=3,6(mol)`
`n_{Cl_2}=n_{MnO_2}=0,9(mol)`
`m_{HCl}=3,6.36,5=131,4(g)`
`V_{Cl_2}=0,9.22,4=20,16(l)`
`b)`
`n_{MnCl_2}=n_{Cl_2}=0,9(mol)`
`m_{dd\ HCl}={131,4}/{20\%}=657(g)`
`C\%_{MnCl_2}={0,9.126}/{657+78,3-0,9.71}.100\%\approx 16,89\%`
`c)`
`Cl_2+2NaOH->NaCl+NaClO+H_2O`
`n_{NaCl}=n_{NaClO}=0,9(mol)`
`n_{NaOH}=2n_{Cl_2}=1,8(mol)`
`C_{M\ NaOH}={1,8}/{0,25}=7,2M`
`C_{M\ NaCl}=C_{M\ NaClO}={0,9}/{0,25}=3,6M`
`d)`
`3Cl_2+2Fe` $\xrightarrow{t^o}$ `2FeCl_3`
`n_{FeCl_3}=2/3n_{Cl_2}=0,6(mol)`
`C\%_{FeCl_3}={0,6.162,5}/{0,6.162,5+52,5}.100\%\approx 65\%`