Consider a tank initially containing 3L of 1M HCl. To this tank, we add 3L of 2M

Question

Consider a tank initially containing 3L of 1M HCl. To this tank, we add 3L of 2M Ca(OH)_2, and allow the acid-base reaction to proceed to completion. At the end of this step, what are the concentrations of HC1, Ca(OH)_2 and CaCl_2 in the mixture ? (Please note that there is water generated during the reaction) We then remove 2L of the contents of the tank, add 2L of 0.5M HC1, and allow the reaction (if any) to proceed to completion what are the concentrations of HC1, Ca(OH)_2 and CaCl_2 in the mixture now ? (Again, note that water may be generated due to the reaction) [NB: You may assume that all of the above occurs at a temperature where the density of water is 1000 kg/m^3 (1000 g/L)]

Explanation / Answer

Ca(OH)2(aq) + 2HCl(aq) ==> CaCl2(aq) + 2H2O(l)

moles of HCl = 3L* 1mol/L = 3 moles

moles of Ca(OH)2 = 3L* 2 mol/L = 6 moles

HCl is limiting reagent. Number of moles of HCl is less than number of moles of Ca(OH)2.

For 1 mol Ca(OH)2 = 2 mol HCl is required

For 0.5 mol Ca(OH)2 = 1 mol HCl is required

For 6 mol Ca(OH)2 = 12 mol HCl is required

moles of Ca(OH)2 in excess = 6-3 = 3 moles

Concentration of Ca(OH)2 = 3M

Concentration of CaCl2 = 3M

Concentration of HCl = 0

(b)

moles of HCl = 2L* 0.5mol/L = 1 mole

moles of Ca(OH)2 = 1L* 2 mol/L = 2 moles

HCl is limiting reagent. Number of moles of HCl is less than number of moles of Ca(OH)2.

For 1 mol Ca(OH)2 = 2 mol HCl is required

For 0.5 mol Ca(OH)2 = 1 mol HCl is required

For 2 mol Ca(OH)2 = 4 mol HCl is required

moles of Ca(OH)2 in excess = 4-2 = 2 moles

Concentration of Ca(OH)2 = 2M

Concentration of CaCl2 = 2M

Concentration of HCl = 0

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---