Question

During a lab experiment performed at STP conditions, you prepare HCl by reacting 100. ml of Cl2 gas with an excess of H2 gas.
How many ml of a solution of Ba(OH)2 0.230M do you need to neutralize all the HCl produced?

Answer 1

Answer:

19.4 mL Ba(OH)2

Explanation:

H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)

At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.

0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2

Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.

0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl

HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.

0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2

Answer 2

The first step in answering this reaction is to write down the required reaction equations and then answer the questions from stoichiometry(mass-mole relationship).

The first equation is;

$H2(g) + Cl2(g) ------->2HCl(g)$

The second equation is

$2HCl(g) + Ba(OH)2(aq) -------> BaCl2(aq) + 2H2O(l)$

From stoichiometry;

1ml = 1cm3

Since hydrogen is in excess, chlorine is the limiting reactant

1 mole of Cl2 occupies 22400cm3

x moles of Cl2 occupies 100cm3

x = 1 * 100/22400

x = 0.00446 moles of Cl2

According to the reaction equation;

1 mole of Cl2 yields 2 moles of HCl

0.00446 moles of Cl2 yields 0.00446 * 2/1

= 0.00892 moles of HCl

From the reaction equation;

2 moles of HCl reacts with 1 mole of Ba(OH)2

0.00892 moles of HCl reacts with 0.00892 * 1/2 = 0.00446 moles of Ba(OH)2

Since 0.00446 moles of Ba(OH)2 reacted

number of moles = concentration * volume

number of moles =  0.00446 moles

concentration = 0.230M

Let the volume in ml be V

0.00446 =  0.230 * V/1000

V = 0.00446 * 1000/ 0.230

V = 19.39 ml

Volume in ml = 19.39 ml of Ba(OH)2

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