Question

How many L of 1.2 M solution of HCl are needed to react completely with 150 mL of 1.6 M HIO3 solution? Assume excess ICI.ICI + HIO3 + 5 HCI → 2 IC13+ 3 H2O0 3.6L02.1L2.5 LO 1.0L

Answer 1

They give us the balanced equation of the reaction. Now, we need to know the moles of HIO3 present in the solution that they describe to us. We are given the molarity and volume of the solution, so the moles will be:

`\begin{gathered} molHIO_3=1.6M*150mL*(1L)/(1000mL) \\ molHIO_3=(1.6molHIO_3)/(1L)*150mL*(1L)/(1,000mL)=0.24molHIO_3 \end{gathered}`

Now, we will find the moles of HCl that will be needed to react 0.24mol of HIO3. We see the stoichiometry of the reaction. The ratio HCl to HIO3 is 5/1. So, the moles of HCl that we need is:

`\begin{gathered} molHCl=GivenmolHIO_3*(5molHCl)/(1molHIO_3) \\ molHCl=0.24molHIO_3*(5molHCl)/(1molHIO_(3))=1.2molHCl \end{gathered}`

They ask us about the volume of the solution, we have the molarity and the moles. So, the volume will be:

`\begin{gathered} Molarity=\frac{MolesSolute}{Lof\text{solution}} \\ LofSolution=(MolesSolute)/(Molarity) \\ LofSolution=(1.2molHCl)/(1.2molHCl/1L)=1.0L \end{gathered}`

To react completely with 150 mL of 1.6 M HIO3 solution are needed 1.0 of 1.2M solution of HCl .

So, the answer will be the last option: 1.0L