I haven\'t finished all of the calculations yet. But I\'m not sure how to find t
ID: 1027341 • Letter: I
Question
I haven't finished all of the calculations yet. But I'm not sure how to find the moles of HCl used. So I would appreciate help on how to solve for the moles of HCl in each trial. The balanced chemical reaction is at the top of the photo by where it says data analysis.
I included this picture because it is the procedure for the lab.
H2COs 20 Data Analysis 2HCVadt Cacus9 coagtHz(20+CaCl2(aa) e the belarced chemica resction equation for the reacian of calium carbonate with hydrocthieric cuce carbon doxide, water and aquecus calcium chionde, What is the mole ratio of calcrum te reacted to carben closide produced? n he number of moles of cakcum cartonate reacted for earh trial. ate the amount of calcium carbonate consumed in the reaction Plot the number of moles of calkcium carbonate in a tablet versus the moles of HCI used. Thirk abaut the piot, wh ich trials will give the most accurate amount of calcium carbonate why? ned in the h poi othe plot that comes Closest to achieving a stochiometric ratio (nether in excess, nether ca theoretical ratio hearenc of hydrochioric acd and calium cartonate. Does the experimental ratio agree with the Data Table Reported Mass a in tablet Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 Trial 7 Trial 8 35. liquid(g) Manzoftablet, aug- 1.-1.IG-1.igla li.my Ilaasll.i R31 Final mass after tablet (35.53 Mass ofou |-.sa 0· lost (g) Males COz last Moles CaCO reacted (mal) Mass of CaCO reacted (g) Moles HCI used (mal) a cid Moles ofExplanation / Answer
Trial 1:
5 mL of 0.5 M of HCl solution has been added. 5 mL = 0.005 L.
Number of moles of HCl = molarity * volume of solution in Litres = 0.005 * 0.5 moles = 0.0025 moles of HCl
Trial 2:
10 mL of 0.5 M of HCl solution has been added. 10 mL = 0.01 L.
Number of moles of HCl = 0.01* 0.5 mole = 0.005 moles
Trial 3:
15 mL of 0.5 M of HCl solution has been added. 15 mL = 0.015 L.
Number of moles of HCl = 0.015* 0.5 mole = 0.0075 moles
Trial 4:
20 mL of 0.5 M of HCl solution has been added. 20 mL = 0.02 L.
Number of moles of HCl = 0.02* 0.5 mole = 0.01 moles
Trial 5:
25 mL of 0.5 M of HCl solution has been added. 25 mL = 0.025 L.
Number of moles of HCl = 0.025* 0.5 mole = 0.0125 moles
Trial 6:
30 mL of 0.5 M of HCl solution has been added. 30 mL = 0.03 L.
Number of moles of HCl = 0.03* 0.5 mole = 0.015 moles
Trial 7:
35 mL of 0.5 M of HCl solution has been added. 35 mL = 0.035 L.
Number of moles of HCl = 0.035* 0.5 mole = 0.0175 moles