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#1
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For those who are experts in Chemistry, right now I'm on a brink of death suffering from my depression for two days for trying to figure out the answer for this problem:
Silver Nitrate reacts with calcium chloride according to the following reaction: 2 AgNO3 (aq) + CaCl2(aq) -- > 2 AgCl(s) + Ca(NO3)2 (aq) Calculate the mass of AgCl that would be formed by reacting 100 cm3 of 0.200 M AgNO3 solution with 100cm3 of 0.500 M CaCl2 solution. Which of the two solutions is the limiting reagent? What will be the concentration of the solution that is in excess? Okay, so I can solve the problem easily if it it had nothing to do with "M" things. I don't udnerstand this. Please help me. Regards. 8) |
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#2
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'M' means Molarity = moles per litre
moles = Molarity x volume (in litres) equation shows a 2:1 ratio of silver nitrate to calcium chloride 1. calculate the moles of 100 cm3 of 0.200 M AgNO3 = 0.02 moles 2. calculate the moles of 100cm3 of 0.500 M CaCl2 solution = 0.05 moles from the equation 0.02 moles of AgNO3 would react with 0.02/2 moles of calcium chloride = 0.01 moles Therefore THERE IS MORE THAN ENOUGH CALCIUM CHLORIDE - it is in excess and the limiting reagent is the silver nitrate. |
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