Calculate the mass of silver chloride that is formed by the interaction of 36 g of hydrochloric acid with a solution of silver nitrate.

Let’s write the reaction equation:

HCl + AgNO3 = AgCl + HNO3;

Now let’s find the number of mol (n) HCl:

n (HCl) = m (HCl): M (HCl) = 36 g: 36.5 g / mol = 0.986 mol (rounded off);

In our equation, before all substances there is conventionally a coefficient of 1, so n for all substances will be equal to n (HCl). Hence n (AgCl) = 0.986 mol.

Now let’s find the mass of AgCl:

m (AgCl) = n (AgCl) * M (AgCl) = 0.986 mol * 143.5 g / mol = 141.491 = 141.5 g (if rounded up).

M (AgCl) = 108 + 35.5 = 143.5 (just add the molar masses of Ag and Cl, which are indicated in the periodic table).