Final answer:
To calculate the mass of precipitate formed, we first need to determine the limiting reactant. From the given information, we can calculate the moles of NaCl and AgNO3. Since NaCl is the limiting reactant, the moles of AgCl formed will be equal to the moles of NaCl. Finally, we convert the mass to milligrams to find that the mass of AgCl formed is 336 mg.
Step-by-step explanation:
To calculate the mass of precipitate formed, we first need to determine the limiting reactant. This can be done by comparing the number of moles of each reactant and seeing which one is present in a smaller amount.
From the given information, we can calculate the moles of NaCl and AgNO3:
moles of NaCl = volume (in L) × molarity
= 0.025 L × 0.094 mol/L
= 0.00235 mol
moles of AgNO3 = volume (in L) × molarity
= 0.015 L × 0.101 mol/L
= 0.00152 mol
Since NaCl is present in a smaller amount, it is the limiting reactant.
The balanced chemical equation tells us that 1 mole of NaCl reacts with 1 mole of AgNO3 to form 1 mole of AgCl. Therefore, the moles of AgCl formed will be equal to the moles of NaCl, which is 0.00235 mol.
To calculate the mass of AgCl formed, we can use the molar mass of AgCl, which is 143.3 g/mol:
mass of AgCl = moles of AgCl × molar mass
= 0.00235 mol × 143.3 g/mol
= 0.336 g.
Finally, we convert the mass to milligrams by multiplying by 1000:
mass of AgCl = 0.336 g × 1000
= 336 mg.