Question

How many grams of butanethiol can be deodorized by reaction with 5.50 mL of 9.80×10−2 M NaOCl?

Answer 1

The reaction occuring between butanethiol and NaOCl-

2C4H10S + NaOCl ---} C8H18S2 + NaCl + H2O

Moles of NaOCl = Molarity*Volume in L

= 9.80*10-2M*0. 0055L

Moles = 0.000539moles

So as per the reaction,

1 mole of NaOCl deodorize 2 moles of butanethiol

So,0.000539moles deodorize =2*0.000539moles=0.001078

Moles of butanethiol = 0.001078moles

Molar mass of butanethiol = 90.1881g/mol

Mass of butanethiol =moles*molar mass

= 0.001078moles*90.1881g/mol

Mass = 0.097g