Question

Q4.23 Unanswered - 2 attempts left Determine the mass of AgCl(s) produced when 0.91 g of AgNO3 aqueous solution reacts with excess Nach aqueous solution as shown below: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Answer 1

Molar mass of AgNO3 = 169.87 g/mol Molar mass of NaCl = 58.44 g/mol Molar mass of AgCl = 143.32 g/mol Balanced chemical equation: 1 AgNO3 + 1 Naci → 1 AgCl + NaNO3 Calculating number of moles: Mass of AgNO3 = 0.91 g Moles = Mass / Molar mass Moles of AgNO3 = ( 0.91 / 169.87 ) Moles of AgNO3 = 0.0054 mol From balanced equation: 1 moles of AgNO3 produces 1 moles of AgCI 0.0054 moles of AgNO3 produces ? moles of AgCI moles of AgCl = ( 1 / 1 ) 0.0054 moles of AgCl = 0.0054 mol Calculating mass: Mass = Moles x Molar mass mass of AgCl = 0.0054 X 143.32 mass of AgCl = 0.76777 g Produced mass of AgCl = 0.77 g