You carefully weigh out 18.00 g of CaCO3 powder and add it to 72.90 g of HCl sol

Question

You carefully weigh out 18.00 g of CaCO3 powder and add it to 72.90 g of HCl solution You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 83.52 g The relevant equation is CaCO: (s) +2HCI(aq) 2O)+CO2(9)+ CaCh(a?) Assuming no other reactions take place, what mass of CO2 was produced in this reaction? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) A. Value Units Submit Previous Answers incorrect; Try Again Next > Provide Feedback

Explanation / Answer

moles CaCO3 = 18 g/100.1 g/mol = 0.18 moles

moles HCl = 72.90 g/36.5 g/mol = 2 moles

limiting reactant = CaCO3

excess moles HCl left after the reaction = (2 - 2 x 0.18)= 1.64 moles

excess mass HCl left after the reaction is done = 1.64 moles x 36.5 g/mol = 59.86 g

mass of solution - mass of exces HCl = 83.520 - 56.86 = 26.66 g

mass fraction CO2 in total products = 44/(44 + 18 + 111) = 0.2543

So,

mass of CO2 produced = 0.2543 x 26.66 = 6.781 g

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