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3 years ago

During a lab experiment performed at STP conditions, you prepare HCl by reacting 100. ml of Cl2 gas with an excess of H2 gas.

Chemistry

1 answer:

Brrunno [24]3 years ago

3 0

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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Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 3.95 M

$V_1$ = volume of stock solution = 268 ml

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Next, you take 100.00 ml of that solution and dilute it to 184 ml in a volumetric flask.

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 1.32 M

$V_1$ = volume of stock solution = 100 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 184 ml

$1.32\times 100=M_2\times 184$

$M_2=0.72M$

Thus concentration of hydrochloric acid in the final solution is 0.72 M

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