Na 2 CO 3 (s) + 2 HCl (aq) 2 NaCl (aq) + H 2 O (l) + CO 2 (g) When 0.235 mols of
ID: 541213 • Letter: N
Question
Na2CO3 (s) + 2 HCl (aq) 2 NaCl (aq) + H2O (l) + CO2 (g)
When 0.235 mols of solid sodium carbonate, Na^CO, (s), is added to 121 mL of 3.67 M hydrochloric acid, HCI(aq), and the carbon dioxide gas, CO, (g), that is formed is collected (Show all of your calculation) B) How many mols of hydrochloric acid. HCI, are present initially? C) If HCl is the limiting reagent, how many moles of CO2 are produced? D) If Na,CO, is the limiting reagent, how many moles of CO, are produced? E) What is the limiting reagent and how many moles of CO2 are produced? F How many moles of carbonate ion, (CO,)2, remain in solution after the rcaction? G) What is the concentration of carbonate ion, [(CO,)2], in molarity remaining in the solution after the reaction? (The final volume of the solution is 125 mL) H) How many mols of sodium ion, Na,remain in solution after the reaction? I) What is the concentration of sodium ion, [Na+, in molarity remaining in the solution after the reaction? (The final volume of the solution is 125 mL.)Explanation / Answer
B) moles of HCl used = 121 mL x 3.67M x10-3 = 0.444moles
C) If HCl were the limiting reagent , the moles of CO2 formed are = moles of HCl /2 (from the balanced eauation)
= 0.444/2
= 0.222 moles
D) If Na2CO3 were the limiting reagent
the moles of CO2 formed = moles of Na2Co3
= 0.235 moles
E) the actual limiting reagent is HCL as 0.235 moles of Na2CO3 require 2x0.235 = 0.470 moles of HCl. But the HCl added is less than 0.470
thus the moles of CO2 formed = 0.222 mol
F) The moles of carbonate remaining = initial moles - reacted moles
= 0.235 - (0.444/2)
= 0.013 moles
G) molarity of carbonate remaining in solution = moles /V(L)
= 0.013 /(125/1000)
=0.104 M
H) the moles of Na+ in solution = 2 x moles of Na2CO3 as one Na2CO3 has two Na + ions and all ions remain in solution as they are spectator ions.
moles of Na+ in solution = 2 x0.235
= 0.470 M
I ) molarity of Na+ ions in solution = moles / volume (L)
= 0.470/0.125
= 3.76
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