What mass of silver chloride can be produced from 1.21 L of a 0.277 M solution of silver nitrate?
To calculate the mass of silver chloride produced, we need to use the stoichiometry of the reaction between silver nitrate and sodium chloride:
AgNO3 + NaCl -> AgCl + NaNO3
From the balanced equation, we can see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to produce 1 mole of silver chloride.
1 L of a 0.277 M solution of silver nitrate contains:
0.277 mol/L x 1.21 L = 0.33517 moles of silver nitrate
Since the molar ratio between silver nitrate and silver chloride is 1:1, we know that 0.33517 moles of silver nitrate will produce 0.33517 moles of silver chloride.
Now, to calculate the mass of silver chloride produced, we need to use the molar mass of silver chloride (AgCl), which is approximately 143.32 g/mol.
0.33517 moles x 143.32 g/mol = 47.97 g
Therefore, 47.97 grams of silver chloride can be produced from 1.21 L of a 0.277 M solution of silver nitrate.