2AgNO_(3)+BaCl_(2)arrow 2AgCl+Ba(NO_(3))_(2) How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride?

To determine how many grams of silver chloride (AgCl) are produced from 5.0 g of silver nitrate (AgNO₃) reacting with an excess of barium chloride (BaCl₂), we need to follow these steps:1. **Write the balanced chemical equation:** 2. **Calculate the molar masses of the reactants and products:** - Molar mass of AgNO₃: \( 107.87 \, (\text{Ag}) + 14.01 \, (\text{N}) + 3 \times 16.00 \, (\text{O}) = 169.87 \, \text{g/mol} \) - Molar mass of AgCl: \( 107.87 \, (\text{Ag}) + 35.45 \, (\text{Cl}) = 143.32 \, \text{g/mol} \)3. **Convert the mass of AgNO₃ to moles:** 4. **Use the stoichiometry of the balanced equation to find the moles of AgCl produced:** According to the balanced equation, 2 moles of AgNO₃ produce 2 moles of AgCl. Therefore, the moles of AgCl produced will be the same as the moles of AgNO₃ used: 5. **Convert the moles of AgCl to grams:** Therefore, 4.21 grams of silver chloride (AgCl) are produced from 5.0 grams of silver nitrate (AgNO₃) reacting with an excess of barium chloride (BaCl₂).