Question

Hydrochloric acid (HCl) reacts with sodium carbonate (Na2C3), forming sodium chloride (NaCl), water (H2O), and carbon dioxide (CO2). This equation is balanced as written:

2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)

Part A: What volume of 1.50 M HCl in liters is needed to react completely (with nothing left over) with 0.250 L of 0.500 M Na2CO3?

Part B: A 621-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 15.1 g CO2. What was the concentration of the HCl solution?

Please show work

Answer 1

A)

find V o fHCl in L

for reaction

mol of Na2CO3 = MV = 0.25*0.5 = 0.125 mol of Na2CO3

ratio is

1 mol of Na2CO3 = 2 mol of HCl

0.125 mol of NA2CO3 = 0.125*2 = 0.250 mol of HCl

M of HCl = mol of HCl / V of HCl

V of Hcl = mol of HCl / M of HCl

V of HCl = 0.250 / 1.5 = 0.16667 Liters required

B)

relate moles of CO2 to HCl

mol of CO2 = mass/MW = 15.1/44 = 0.343181 mol of CO2

relate CO2 to HCl

2 mol of HCl = 1 mol of CO2

0.343181 mol of CO2 = 2*0.343181 = 0.686362 mol of HCl

now...

[HCl] = mol of HCl / V of HCl = 0.686362 / (0.621) = 1.105 M of HCl