Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO3 is ad

Question

Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO3 is added to 0.175 L of 0.450 M CaCl2. The equation is: 2 AgNO3(aq)+CaCl2(aq) -->2 AgCl(s)+Ca(NO3)2(aq)

Explanation / Answer

AgNO3 + CaCl2 --> AgCl + Ca(NO3)2 don't forget ALL of the products. you have .2 L of 0.2 M, molarity is moles/L, so 0.2 M = moles/.2 L or 0.04 moles AgNO3 lly for CaCl2 moles = 0.175*0.45 = 0.07875 moles of CaCl2 now, balance the equation. 2AgNO3 + CaCl2 --> 2AgCl + Ca(NO3)2 2 moles AgNO3 produces 2 moles AgCl 1 mole CaCl2 produces 2 moles AgCl which is your limiting reactant, or the reactant that produces the least amount of product. 0.04 moles AgNO3 would produce 0.04 moles AgCl 0.07875 moles CaCl2 would produce 0.1575 moles AgCl therefore, AgNO3 is your limiting reactant and this is all that you can produce. now, convert moles AgCl to grams by multiplying by the molar mass of AgCl ans : 0.04*143.32 g = 5.7328 g

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