Question

What weight of $AgCl$ would be precipitated if $10ml$ of $HCl$ gas measured at $12^\circ C$ and $750mm$ pressure were passed into the excess of a solution of silver nitrate?

Answer 1

When hydrochloric acid $HCl$ reacts with silver nitrate it forms a white precipitate of silver chloride. It is a double displacement reaction because the hydrogen atom acquires the position of silver whereas silver acquires the position of hydrogen forming silver chloride.

Complete answer:
The first thing to do is to write down the reaction that is described in the question. Since the reactant and one of the product are given, we can write the balanced reaction:
$AgN{O_3} + HCl \to AgCl + HN{O_3}$
In this type of question. It is nice to write down all the details and information we are provided with so we can have a clear look at what to do.
The given things are:
Pressure = $750mm$ $= \dfrac{{750}}{{760}}atm$
Volume= $10ml$ $= 10 \times {10^{ - 3}}L$
Temperature= $12^\circ C$ $= 12 + 273K = 285K$
We will calculate the number of moles of $HCl$ through the ideal gas equation that is $PV = nRT$
$10 \times {10^{ - 3}} \times \dfrac{{750}}{{760}} = n \times 0.082057 \times 285$ (Where R is gas constant)
$n = \dfrac{{750 \times {{10}^{ - 3}}}}{{760 \times 0.082057 \times 285}} = 0.042 \times {10^{ - 2}}$
Therefore, moles of $HCl$ is $0.042 \times {10^{ - 2}}$
Now we will calculate the molecular mass of $AgCl$ $= 1 \times Ag + 1 \times Cl = 107.9u + 35.45u = 143.35u$
Number of moles of $AgCl$ = number of moles of $HCl$
$\dfrac{w}{{143.5}} \times 1000 = 0.422$
$w = \dfrac{{0.422 \times 143.5}}{{1000}} = 0.605g$

Note:
For this particular question, it seems useless to calculate the molar mass of the other reactant, but for future questions, they can ask more about the question than we need to calculate the molar mass of other reactants. All the other factors should be in their S.I unit. Example- pressure, temperature, mass of solute or mass of solvent, volume etc.