Use stoichiometry to back calculate from 5.3 g CaCl2 to see the moles HCl used (you could go the long way and calculate moles CaCO3 used then convert to HCl but that step isn't necessary).
moles CaCl2 = grams/molar mass = 5.3/about 111 = about 0.048 (but you need to do it again more accurately).
Then convert moles CaCl2 to moles HCl using the coefficients in the balanced equation. 0.048 moles CaCl2 x (2 moles HCl/1 mole CaCl2) = 0.048 x (2/1) = about 0.096 mols HCl.
Then the definition of molarity is M = moles/L; therefore, M HCl = moles HCl/L HCl. You have moles from above, L HCl is given in the problem (25 mL = 0.025L), solve for M.
What is the molarity of the hydrochloric acid solution used if 25mL of it reacted with excess calcium carbonate to produce 5.3g of calcium chloride?
CaCo3(s) + 2HCl(aq)---> CaCl2(aq) + CO2(g) + H2O(l)
I am really confused on even what this question is asking..any help would be greatly appreciated.
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