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A chemistry laboratory is provided with a 10 mL solution containing a mixture of sodium chloride and potassium chloride. After adding an excess of silver nitrate, it has been observed that 0.476 grams of AgCl precipitate out. Determine the mass of each component (NaCl and KCl) in the original solution using mass balance calculations.

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To determine the mass of each component  NaCl and KCl  in the original solution, we will first need to find the moles of AgCl precipitated and then use stoichiometry to find the moles of NaCl and KCl. Finally, we will convert the moles of NaCl and KCl back to mass.1. Calculate the moles of AgCl precipitated:AgCl has a molar mass of 143.32 g/mol  107.87 g/mol for Ag + 35.45 g/mol for Cl .0.476 g AgCl *  1 mol AgCl / 143.32 g AgCl  = 0.00332 mol AgCl2. Use stoichiometry to find the moles of NaCl and KCl:The reaction between NaCl and AgNO is:NaCl + AgNO  NaNO + AgClThe reaction between KCl and AgNO is:KCl + AgNO  KNO + AgClSince both reactions produce 1 mol of AgCl for each mole of NaCl or KCl, the total moles of NaCl and KCl in the original solution is equal to the moles of AgCl precipitated:0.00332 mol NaCl + 0.00332 mol KCl = 0.00332 mol AgClLet x be the moles of NaCl and y be the moles of KCl. Then,x + y = 0.003323. Convert the moles of NaCl and KCl back to mass:Since we don't have enough information to determine the exact ratio of NaCl to KCl in the original solution, we can only express the mass of each component as a function of the other.Mass of NaCl = x *  58.44 g/mol  = 58.44x gMass of KCl = y *  74.55 g/mol  = 74.55y gWe can rewrite the equation from step 2 in terms of mass:58.44x + 74.55y = 0.00332 *  58.44 + 74.55  = 0.441 gThis equation represents the mass balance of NaCl and KCl in the original solution. Without additional information, we cannot determine the exact mass of each component. However, we can say that the sum of the masses of NaCl and KCl in the original solution is 0.441 g.

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