Consider the following equation:
Caco3(s) + HCl(l) --> CaCl(s) + HCO3(l)
there is 500mL of Hydrochloric acid initially.
-calculate the mass of the calcium carbonate which would have reacted.
-calculate the initial concentration (molarity) of the HCl
2007-05-22 03:37:15 · 4 answers · asked by hiscageofstars 1 in Science & Mathematics ➔ Chemistry
there is a loss in mass of the reactants of 14grams.
2007-05-22 03:56:13 · update #1
If we consider the complete chemical reaction it sums up as follows
CaCO3 + 2HCl ----> CaCl2 + H2O + CO2.
Among the products the only gas is CO2. So it will be marked as 'loss in reactant' . 14 grams corresponds to 14/44 = 0.318 moles.
One mole of CO2 is produced for every 2 moles of HCl and 1 mol of calcium carbonate consumed. So the mass of calcium carbonate = 0.318*100=31.8 grams.
Moles of HCl is 0.636 in 1/2 L. In one L it is 1.2. So molarity is approximately 1.2 L.
2007-05-22 04:59:16 · answer #1 · answered by Ajinkya N 5 · 0⤊ 0⤋
Impossible to solve. You must give at least, the density of HCl liquid
I found in wilkipedia : 1.18g/cm^3
if this is correct so 500ml = 500*1.18=590g
this corresponds to 590/36.46 (36.46 = molecular weight) =16.18mole
in the equation you see that 1 mole of CaCO3 reacts with 1 mole of HCl. so you need 16.18 mole of CaCO3
the mass of 1 mole of CaCO3 = 40+12+3*16 =100g
and the mass CaCO3 you search =100*16.18=1618g
molarity of HCl 16.18 in 0.5L
molariy is 16.18/0.5= 32.36M
2007-05-22 03:55:14 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
Impossible to calculate. 500ml of HCl - what is the concentration?
How much CaCO3 is there?
2007-05-22 03:45:09 · answer #3 · answered by Dr Dave P 7 · 0⤊ 0⤋
sir i think u did not provide sufficient info for us to calculate the answer for u
2007-05-22 03:51:01 · answer #4 · answered by Anonymous · 0⤊ 0⤋