Final answer:
To find the number of grams of AgCl produced from 30.0 grams of CaCl2, we need to use stoichiometry. The correct answer is 77.5 grams.
Step-by-step explanation:
To find the number of grams of AgCl produced from 30.0 grams of CaCl2, we need to use stoichiometry. First, calculate the number of moles of CaCl2 using its molar mass:
Moles of CaCl2 = mass / molar mass = 30.0 g / (40.078 g/mol + 2 * 35.453 g/mol) = 0.371 moles
According to the balanced equation, the molar ratio between CaCl2 and AgCl is 1:2. Therefore, the moles of AgCl produced will be twice the moles of CaCl2:
Moles of AgCl = 2 * 0.371 moles = 0.742 moles
Finally, calculate the mass of AgCl:
Mass of AgCl = moles * molar mass = 0.742 moles * (107.868 g/mol + 35.453 g/mol) = 77.5 grams