2 Shitt Alt Alt Ctrl Name Gases #5 Date A dry 2.50-gram mixture consisting of so
ID: 1041410 • Letter: 2
Question
2 Shitt Alt Alt Ctrl Name Gases #5 Date A dry 2.50-gram mixture consisting of sodium hydroxide, sodium carbonate, and sodium chloride is made to react with 0.50 L of excess 1.0 M HCI solution. culate the percentage of sodium carbonate in the mixture if 125-mLof dry CO gas, measured at 22°C and 740 torr, is obtained from the reaction. The excess HCl is found by titration to be chemically equivalent to 43.5 mL of 1.20 M KOH. What percentages of sodium hydroxide and of sodium chloride in the original mixture. (Pwaer at 22.0°C is 19.8 torr) b.Explanation / Answer
Sodium hydroxide (NaOH) and sodium carbonate (CO2) will react with hydrochloric acid (HCl) as below.
NaOH (aq) + HCl (aq) --------> NaCl (aq) + H2O (l)
Na2CO3 (aq) + 2 HCl (aq) --------> 2 NaCl (aq) + CO2 (g) + H2O (l)
As per the stoichiometric equations above,
1 mole NaOH = 1 mole HCl.
1 mole Na2CO3 = 2 moles HCl = 1 mole CO2.
(a) 125 mL of dry CO2 gas was collected at 740 torr and 22°C. The vapor pressure of water at 22°C is 19.8 torr.
Therefore, the pressure of dry CO2 is P = (740 – 19.8) torr = 720.2 torr = (720.2 torr)*(1 atm/760 torr) = 0.9476 atm.
The volume of the gas is V = 125 mL = (125 mL)*(1 L/1000 mL) = 0.125 L while the temperature of CO2 is T = 22°C = (22 + 273) K = 295 K.
Use the ideal gas law
P*V = n*R*T where n = number of moles of CO2
=====> (0.9476 atm)*(0.125 L) = n*(0.082 L-atm/mol.K)*(295 K)
=====> n = (0.9476 atm)*(0.125 L)/(0.082*295 L-atm/mol) = 0.004897 mole.
As per the above stoichiometric equations,
0.004897 mole CO2 = 0.004897 mole Na2CO3.
Molar mass of Na2CO3 = (2*22.989 + 1*12.011 + 3*15.999) g/mol = 105.986 g/mol.
Mass of Na2CO3 in the mixture = (0.004897 mole)*(105.986 g/mol) = 0.5190 g ? 0.52 g.
Percentage Na2CO3 in the mixture = (0.52 g)/(2.50 g)*100 = 20.80% (ans).
Please check the values, because I am getting an absurd result for part (b).
(b) Moles of HCl added initially = (0.50 L)*(1.0 M) = 0.50 mole.
Moles of HCl reacted with 0.004897 mole Na2CO3 = (0.004897 mole Na2CO3)*(2 moles HCl/1 mole Na2CO3) = 0.009794 mole.
Moles HCl left = (0.50 – 0.009794) mole = 0.490206 mole.
HCl reacts with KOH as per the equation,
KOH (aq) + HCl (aq) -------> KCl (aq) + H2O (l)
As per the stoichiometric equation,
1 mole KOH = 1 mole HCl.
Therefore, moles KOH = moles HCl neutralized = (43.5 mL)*(1 L/1000 mL)*(1.20 M) = 0.0522 mole.
Moles HCl reacted with NaOH = 0.490206 – 0.0522 = 0.438006 mole.
Moles NaOH in the mixture = moles HCl reacted = 0.438006 mole.
Molar mass of NaOH = (1*22.989 + 1*15.999 + 1*1.008) g/mol = 39.996 g/mol.
Mass of NaOH in the mixture = (0.438006 mole)*(39.996 g/mol) = 17.518 g ? 17.52 g.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.