Final answer:
To determine the mass of silver chloride precipitating from a reaction between silver nitrate and barium chloride, calculate the moles of BaCl2 in the solution, double that to find the moles of AgCl, and then convert to mass giving 171.984 grams AgCl.
Step-by-step explanation:
The student asked how many grams of silver chloride (AgCl) would precipitate from the reaction between silver nitrate and a 0.400 molar solution of barium chloride (BaCl2) given a volume of 1500 mL. To find the answer, we'll first need to determine the number of moles of barium chloride in the solution using the molarity formula: moles = molarity × volume (in liters). The reaction between silver nitrate and barium chloride can be represented by the balanced chemical equation:
2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)
According to the equation, two moles of silver chloride are produced for every mole of barium chloride. After calculating the number of moles of BaCl2, we simply double that amount to get the moles of AgCl, and then convert that to grams using the molar mass of AgCl. Here are the calculations:
- Moles of BaCl2 = 0.400 mol/L × 1.5 L = 0.600 mol
- Moles of AgCl = 2 × 0.600 mol = 1.200 mol
- Mass of AgCl = 1.200 mol × 143.32 g/mol (molar mass of AgCl) = 171.984 grams
171.984 grams of silver chloride is the mass that would precipitate from the given solution.