Hydrochloric acid (HCl) reacts with sodium carbonate (Na_2CO_3), forming sodium

Question

Hydrochloric acid (HCl) reacts with sodium carbonate (Na_2CO_3), forming sodium chloride (NaCl), water (H_2O), and carbon dioxide (CO_2). This equation is balanced as written: 2HCl(aq) + Na_2CO_3(aq) rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g) What volume of 2.25 MHCl in liters is needed to react completely (with nothing left over) with 0.250 L of 0.100 MNa_2CO_3? Express your answer to three significant figures and include the appropriate units. A 681-mL sample of unknown HCl solution reacts completely with Na_2CO_3 to form 14.1 g CO_2 What was the concentration of the HCl solution? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A)

mol of Na2CO3 = MV = 0.25*0.1 = 0.025 mol of Na2CO3

ratio is 2:1

so

0.025 mol of Na2CO3 --> 2*0.025 = 0.050 mol of HCl

for volume of acid

M = mol/V

V = mol/M = 0.050 /2.25 = 0.02222 Liters = 0.02222*10^3 mL = 22.22 mL

B)

find mol of CO2

MW = 44 g/mol

mol = mass/MW = 14.1/44 = 0.320454 mol of CO2 formed

so...

ratio is

2 mol of HCl --> 1 mol of CO2

mol of HCl = 0.320454*2 = 0.640908 mol of HCl

then

M = mol/V = (0.640908)/(681*10^-3) = 0.94112 M

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