Question

what volume (in L) of 4.35 M CHI solutions is needed to react with 18.5 g of Ca(OH)2

Answer 1

For a reaction to be completed, the two reagents should have the same amount of moles:

$n(HI) = n(Ca(OH)_2)$

$n(Ca(OH)_2) = \frac{m}{M} = \frac{18.5g}{74.1 g/mol} = 0.25 mol$

The mole number of HI is 0.25 mol according to the first equation and can be written in the following way:

$n(HI) = c\times V$

Now the volume is:

$V(HI)=\frac{n}{c}=\frac{0.25 mol}{4.35 mol/l} = 0.057 L$