Question

1. 200.0mL of hydrochloric acid solution reacted with 200.0 mL of sodium carbonate solution used as an excess reactant. 

227.0mL of carbon dioxide gas was collected at 22.0OC temperature and 1.00 Atm pressure. Yield was 85%

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

               What was the concentration of hydrochloric acid?

Answer 1

Ideal gas low:

pV=nRT

V(CO₂)=227.0 mL= 0.227 L

p=1 atm

T= 22.0 °C = 22+273=295 K

R = 0.08206 L×atm/mol×K

$n= \frac{pV}{RT} \\ n(CO_2) = \frac{1 \times 0.227}{0.08206 \times 295}= 0.009377 \;mol$

Proportion according to the yield:

0.009377 – 85%

x – 100%

$x= \frac{0.009377 \times 100}{85}= 0.01103 \;mol$

According to the reaction equation:

n(HCl) = 2n(CO₂) $= 2 \times 0.01103 = 0.02206 \;mol$

Concentration of HCl:

$C= \frac{n}{V}$

V(HCl) = 200.0 mL = 0.2 L

$C(HCl)= \frac{0.02206}{0.2} = 0.1103 \;mol/L$

Answer: 0.1103 M