Answer:
Q5.
a} Yes , All Cl2 get eliminated by adding 170 g of sodium thiosulfate .
b) There excess of Sodium thiosulfate by 2.637 g
Q6.
a) Mass = 162 g Sodium bicarbonate
b)Volume of Ammonia remain after the reaction = 0.086 x 24.8 = 1.60 L
Explanation:
Points to be considered :
There is a difference between STP and SATP :
STP = Standard Temperature and Pressure (273.15 K and 1 atm)
1 mole of gas at STP = 22.4 L
SATP = Standard Ambient Temperature and Pressure (293.15 K and 1 atm)
1 mole of gas at SATP = 24.8 L
[tex]moles = \frac{Given\ mass}{Molar\ mass}[/tex]
Number of moles of gas at SATP :
[tex]moles = \frac{Given\ Volume}{24.8L}[/tex]
Q5.
First, calculate the number of moles of Cl2 and thiosulfate present in the reaction :
Volume of Cl2 = 105 L
Moles of Cl2 =
[tex]moles = \frac{Given\ Volume}{24.8}[/tex]
[tex]moles = \frac{105}{24.8}[/tex]
Moles of Cl2 in Reaction medium = 4.2338 mole
Mass of Sodium thiosulfate = 170 g
Molar mass of thiosulfate = 158.11 g/mol (theoretical value)
[tex]moles = \frac{170}{158.11}[/tex]
= 1.075
Moles of Sodium Thiosulfate in Reaction medium = 1.075 mole
To check whether the given moles of Cl2 and sodium thiosulfate satisfy theoretical values :
Consider the Given reaction and apply law of conservation of mass
[tex]4Cl_{2}+Na_{2}S_{2}O_{3}+5H_{2}O\rightarrow 2NaHSO_{3}+8HCl[/tex]
[tex]Na_{2}S_{2}O_{3}[/tex] = sodium thiosulfate
This equation indicates ,
4 moles of Cl2 require = 1 mole of sodium thiosulfate
1 mole of Cl2 require =
[tex]\frac{1}{4}[/tex] of sodium thiosulfate = 0.25
4.2338 mole of Cl2 should need = 0.25 x 4.2338
= 1.058 mole of sodium thiosulfate
Required Thiosulfate = 1.058 mole
But,
Moles of Sodium Thiosulfate in Reaction medium = 1.075 mole
So , extra moles of Sodium Thiosulfate is present in the reaction by
= 1.075 - 1.0589 = 0.0166 mol
Molar mass of sodium thiosulfate = 158 .11 g/mol
[tex]Mass =0.166\times 158.11[/tex]
= 2.637 g
a} .Yes , All Cl2 get eliminated by adding 170 g of sodium thiosulfate .
b) There excess of Sodium thiosulfate by 2.637 g
Q6.
Volume of ammonia = 50.0 L
Moles of Ammonia ,
[tex]moles = \frac{Given\ Volume}{24.8L}[/tex]
[tex]moles = \frac{50}{24.8L}[/tex]
= 2.016 moles
Moles of CO2 =
Mass of CO2 = 85.0 g
Molar mass = 44 g/mol
[tex]moles = \frac{Given\ mass}{Molar\ mass}[/tex]
[tex]moles = \frac{85.0}{44}[/tex]
= 1.932 mol of CO2
Now check the law of conservation of mass :
[tex]NaCl + NH_{3} + CO_{2} +H_{2}O\rightarrow NaHCO_{3} +NH_{4}Cl[/tex]
According to above equation ,
1 mole of CO2 Needs = 1 mole of NH3
1.93 mol of CO2 need = (1 x 1.93) mol
1.93 mol of CO2 need = 1.93 mol of ammonia
Available ammonia = 2.016 mol
So Ammonia is in excess by:
= 2.016 - 1.93 mol
= 0.086 mol
Volume at SATP is calculated by
[tex] V =mole\times 24.8[/tex]
Volume of Ammonia remain after the reaction = 0.086 x 24.8
= 2.1104 L
= 2.10 L
CO2 is the limiting reagent and governs the product formation :
Molar mass of NaHCO3 = 84.007 g/mol
1 mole of CO2 Needs = 1 mole of NaHCO3 = 84.007
1.93 mol of CO2 need = 1.93 x 84.007 mol
= 162.133 g of Sodium bicarbonate
= 162 g Sodium bicarbonate
Note : The answers are present in rounded figures .
