Question

Can someone help on five and six I am at a loss on how to solve it

Answer 1

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

Step-by-step 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

`moles = (Given\ mass)/(Molar\ mass)`

Number of moles of gas at SATP :

`moles = (Given\ Volume)/(24.8L)`

Q5.

First, calculate the number of moles of Cl2 and thiosulfate present in the reaction :

Volume of Cl2 = 105 L

Moles of Cl2 =

`moles = (Given\ Volume)/(24.8)`

`moles = (105)/(24.8)`

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)

`moles = (170)/(158.11)`

= 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

`4Cl_(2)+Na_(2)S_(2)O_(3)+5H_(2)O\rightarrow 2NaHSO_(3)+8HCl`

`Na_(2)S_(2)O_(3)` = sodium thiosulfate

This equation indicates ,

4 moles of Cl2 require = 1 mole of sodium thiosulfate

1 mole of Cl2 require =

`(1)/(4)` 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

`Mass =0.166* 158.11`

= 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 ,

`moles = (Given\ Volume)/(24.8L)`

`moles = (50)/(24.8L)`

= 2.016 moles

Moles of CO2 =

Mass of CO2 = 85.0 g

Molar mass = 44 g/mol

`moles = (Given\ mass)/(Molar\ mass)`

`moles = (85.0)/(44)`

= 1.932 mol of CO2

Now check the law of conservation of mass :

`NaCl + NH_(3) + CO_(2) +H_(2)O\rightarrow NaHCO_(3) +NH_(4)Cl`

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

`V =mole* 24.8`

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 .