What mass of silver chloride can be produced from 1.27 L of a 0.277 M solution of silver nitrate?
To solve this problem, we can use the stoichiometry of the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) to calculate the mass of silver chloride (AgCl) that can be produced.
The balanced chemical equation for this reaction is:
AgNO3 + NaCl -> AgCl + NaNO3
From this equation, we see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.
First, we need to calculate the number of moles of AgNO3 in 1.27 L of a 0.277 M solution:
mol = Molarity x Volume
mol = 0.277 mol/L x 1.27 L
mol = 0.35219 mol
Since the molar ratio between AgNO3 and AgCl is 1:1, the number of moles of AgCl produced will be the same as the number of moles of AgNO3:
0.35219 mol of AgCl will be produced.
Next, we need to calculate the molar mass of AgCl, which is 143.32 g/mol.
Finally, we can calculate the mass of AgCl produced:
mass = moles x molar mass
mass = 0.35219 mol x 143.32 g/mol
mass = 50.48 g
Therefore, 50.48 grams of silver chloride can be produced from 1.27 L of a 0.277 M solution of silver nitrate.