Question

`(i)` What volume of `5.00*mol*L^-1` `HCl(aq)` is required to react with a `17.0*g` mass of sodium hydroxide. `(ii)` what volume of sodium sulfide at `0.100*mol*L^-1`, is required to precipitate a `33.0*mL` volume of `0.100*mol*L^-1` of `AgNO_3`?

Answer 1

We need to find (i) write a stoichiometrically balanced equation, and (ii) we need to find the molar quantities of each reagent.

Explanation:

Our basic equation is......

`"Concentration"="Moles of solute"/"Volume of solution"`.

And thus by taking a product or a quotient, we can get the number of moles or the volume of solution.

`NaOH(s) + HCl(aq) rarr NaCl(aq) + H_2O(l)`

And thus, for the first equation there is 1:1 stoichiometry. And moles of `NaOH` are equivalent to moles of `HCl`.

`"Moles of NaOH"-=(17.0*g)/(40.0*g*mol^-1)=0.425*mol`

And by the give stoichiometry, we need ONE equiv of `HCl(aq)`.

We are given `5.00*mol*L^-1` `HCl(aq)`, and thus we solve the quotient......

`(0.425*cancel(mol))/(5.00*cancel(mol*L^-1))xx10^3*mL*cancel(L^-1)=85.0*mL`.

Clear?

For the second scenario, you must simply KNOW that `Ag_2S` (like most sulfides save those of the alkali metals) is INSOLUBLE.

`2AgNO_3(aq) + Na_2S(aq) rarr Ag_2S(s)darr+2NaNO_3(aq)`

The net ionic equation is simply......

`2Ag^+ + S^(2-) rarr Ag_2S(s)darr`

`"Moles of"` `AgNO_3=33.0xx10^-3*Lxx0.100*mol*L^-1=`
`3.30xx10^-3*mol`

And thus we need half an equiv of `Na_2S`, i.e.

`(1/2xx3.3xx10^-3*mol)/(0.100*mol*L^-1)xx10^3*mL*L^-1`

`"under 17 mL"` of the sodium sulfide solution.........