Question

# Hydrochloric acid (HCl) reacts with sodium carbonate (Na2C3), forming sodium chloride (NaCl), water (H2O), and carbon...

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?

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

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