Hydrochloric acid (HCl) reacts with sodium carbonate (Na2CO3), forming sodium ch

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 2.50 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.500 M Na2CO3? Express your answer to three significant figures and include the appropriate units. Part B A 651-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 12.1 g CO2. What was the concentration of the HCl solution? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

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

Part A: We can simply proceed by equating the moles of reactants on the two sides of the equation.

Let volume of HCl needed be x L.

Then,

Moles of Na2Co3=Moles of HCl

0.500M * 0.750 L = 2 * 2.50M * x L

On solving, x=0.075 L

Part B.

In this part, like in part A, we equate the moles on the two sides of the equation,

Let the concentration of HCl be xM.

Moles of HCl= Moles of Na2Co3

12.1g / 44 g/Mol= 2 * 0.651 L * x M

On solving, x= 0.037M

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