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Not a timed or graded assignment. Please show calculation work. Question prompt : “what’s the limiting reactant”. Quick answer = amazing review :)

Not a timed or graded assignment. Please show calculation work. Question prompt : “what-example-1
User TJHeuvel
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1 Answer

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In order to find the limiting and excess reactant in a question, the first thing we need to do is set up the properly balanced equation:

CaCO3 + 2 HCl -> CaCl2 + H2O + CO2

Now we know that the molar ratio between CaCO3 and HCl is 1:2, this will be important in order to determine the limiting and excess reactant, and we also have:

10.0 grams of CaCO3, molar mass 100.086g/mol

15.0 grams of HCl, molar mass 36.46g/mol

Now we need to find out how many moles of reactant we have and see if by their molar ratio, these values are proportional

CaCO3:

100.086g = 1 mol

10.0g = x moles

x = 0.099 moles of CaCO3 in 10 grams

By the molar ratio, we should have 2 times more moles of HCl than CaCO3, and 0.099*2 = 0.198 moles of HCl, let's check if that is what we actually have

HCl:

36.46g = 1 mol

15.0g = x moles

x = 0.411 moles, this value is way more than we actually need, making it HCl the excess reactant and CaCO3 will be the limiting reactant

User Cocorossello
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