Sodium chorate, `NaClO_(3)`, can be prepared by the following series of reactions:
`2KMnO_(4)+16HCl rarr 2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)`
`6Cl_(2)+6Ca(OH)_(2)rarr Ca(ClO_(3))_(2)+5CaCl_(2)+6H_(2)O`
`Ca(ClO_(3))_(2)+Na_(2)SO_(4) rarr CaSO_(4)+2NaClO_(3)`
What mass of `NaClO_(3)` can be prepared from `100 mL` of concentarted HCl (density `1.18 gm//mL` and `36%` by mass)? Assume all other substance are present in excess amounts.

Explanation

`5 "mole" Cl_(2)` obrained from `= 16 "mole"HCl`
`6"mole" Cl_(2)` is obtained from `= (16)/(5)xx6=(96)/(5)"mole"HCl`
`6"mole"Cl_(2)` or `(96)/(5)"mole" HCl` gives `=2 "mole" NaCl_(3)`
`(42.48)/(36.5)"mole"HCl` gives `=(2)/(96)xx5xx(42.48)/(36.5)"mole" NaCl_(3)`
`=(10)/(96)xx(42.48)/(36.5)xx106.5= 12.9 g`