Answer :
To find the mass of silver chloride produced from 1.55 L of a 0.183 M solution of silver nitrate, first calculate the moles of AgNO3 and then multiply by the molar mass of AgCl, resulting in 40.67 grams of AgCl.
To find the mass of silver chloride that can be produced from 1.55 L of a 0.183 M solution of silver nitrate, we need to perform stoichiometric calculations. The chemical reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) to form silver chloride (AgCl) and sodium nitrate (NaNO3) is as follows:
AgNO3 + NaCl ⇒ AgCl + NaNO3
This reaction has a stoichiometry of 1:1 between silver nitrate and silver chloride. First, calculate the moles of AgNO3 in the solution:
Moles of AgNO3 = Volume (L) x Molarity (M)
= 1.55 L x 0.183 M
We then use the molar mass of AgCl to convert moles of AgNO3 to grams of AgCl:
Mass of AgCl = Moles of AgNO3 x Molar mass of AgCl
Finally, carry out these calculations to find the mass of AgCl produced:
Moles of AgNO3 = 1.55 x 0.183
= 0.28365 mol
Mass of AgCl (given that the molar mass of AgCl is 143.32 g/mol) = 0.28365 mol x 143.32 g/mol
= 40.67 grams of AgCl
