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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?

Sagot :

Mitak5575idn

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

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