If 30 ml of 0.150M cacl is added to 40ml of 0.1 M AgNo3. What is the mass of AgC

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Explanation / Answer

CaCl2(aq) + 2AgNO3(aq) ----> 2AgCl(s) + Ca(NO3)2)aq) First work out moles of each reactant that you added. Molarity = moles / litres therefore moles = Molarity x Litres Number of moles in 70.0 mL (0.0700 L) of 0.150 M solution = Molarity x Litres = 0.150 M x 0.0700 L = 0.0105 moles od CaCl2 Number of moles of AgNO3 in 15.0 ml (0.0150 L) 0.100 M solution = 0.100 M x 0.0150 L = 0.00150 moles of AgNO3 Now you know the moles of each reactant you can workl out the limiting reagent. The limiting reagent is the reagent which is not provided in enough quantity to fully react with all of the other reagent. There are 0.00150 moles of AgNO3 The equation states that 2 moles of AgNO3 are required to react with 1 mole of CaCl2 So 1 moles AgNO3 requires 1/2 moles of CaCl2 Therefore 0.00150 moles AgNO3 needs 1/2 x 0.00150 moles CaCl2 = 0.000750 moles of CaCl2 are required to react with all the AgNO3 We have 0.0105 moles of CaCl2, therefore we have more CaCl2 then we need. Therefore The AgNO3 is the limiting reagent. The maximum amount of product you can form is from full reaction of the limiting reagent, in this case 0.00150 moles of AgNO3 2 moles of AgNO3 react to form 2 moles of AgCl therefore 0.00150 moles will form 0.00150 moles of AgCl Now work out mass of 0.00150 moles AgCl moles = mass / molar mass therefore mass = molar mass x moles molar mass AgCl = 107.9 + 35.45 = 143.35 g/mol mass = 143.35 g/mol x 0.00150 moles = 0.215 g of AgCl

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