Question

Hydrochloric acid (HCl) reacts with sodium carbonate (Na2CO3), 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 2.75 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.400 M Na2CO3?

Part B: A 675-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 18.1 g CO2. What was the concentration of the HCl solution?

Answer 1

the reaction is:

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

from the stereochemical coefficients of the reaction, it is clear that 2 moles of Hcl reacts with one mole of Na2CO3 to give 2 moles of NaCl, 1 moles of H2O and 1 moles of CO2.

part A:

now lets calculate the moles of Na2CO3

moles = C x V
0.400 mol/L x 0.750 L = 0.3 moles

now we will Use the mole ratio between HCl and Na2CO3 from the equation to find the number of moles of HCl needed to completely use up all the Na2CO3. Can you see that for every mole of Na2CO3 you need twice that many moles of HCl.

So moles of HCl will be 0.3 moles x 2/1 = 0.60 moles of HCl

Now that we have both moles and Concentration of HCl we can find the volume.

C = n / V , so V = n / C

V = 0.6 mol / 2.75 mol/L = 0.218 L = 218 mL

part B:

let the concentration = c

moles of CO2 formed:

molar mass= 44 g

moles= 18.1/44 = 0.411 moles

2 moles of HCl reacts to form 1 moles of CO2.

So moles of HCl= 2x0.411 = 0.822 moles

c=n/V

c= 0.822/0.675 = 1.22 M