Answer : The correct answer is 1) AlCl₃ - CH₃Cl 2) HNO₃ -H₂SO₄ at room temperature 3) Fuming HNO₃ -H₂SO₄ at 90-100 ⁰ C heat .
I think this reaction is forming 2,4,6- trinitrotoluene from benzene, since the product is not mentioned. Following are the steps to convert Benzene to 2,4,6 trinitrotoluene .
Step 1: Conversion of Benzene to Toluene .
Benzene can be converted to toluene by Friedel Craft Alkylation of benzene . In this reaction reagent AlCl₃ and Ch3Cl is used . Electrophile CH³⁺ is produced which attached on carbon of benzene and formation of Toluene and HCl occur.
![Benzene \frac{AlCl3}{Ch3Cl}> Toulene + HCl](https://tex.z-dn.net/?f=%20Benzene%20%20%20%20%20%5Cfrac%7BAlCl3%7D%7BCh3Cl%7D%3E%20%20%20Toulene%20%2B%20HCl%20)
Step 2 : Conversion of Toluene to dinitrotoluene.
Dinitritoluene is prepared from toluene by Nitration . This reaction uses Electrophilic substitution mechanism . The reagents used are HNO₃ and H₂SO₄ at room temperature . These reagents produces NO₂⁺ ( nitronium ion ), a electrophile which attacks on C2 and C4 Carbon atoms of Toluene.
Toluene ![Tolune \frac{HNO3 -H2SO4}{30-40 degree C} -> 2,4- dinitrotoluene](https://tex.z-dn.net/?f=%20Tolune%20%20%20%5Cfrac%7BHNO3%20-H2SO4%7D%7B30-40%20degree%20C%7D%20-%3E%20%202%2C4-%20dinitrotoluene%20)
Step 3) Conversion of Dinitro toluene to trinitrotoluene.
This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.
Dinitrotoluene ![2,4 -dinitrotoluene \frac{fuming HNO3-H2So4}{90-100 C} -> 2,4,6-trinitrotoluene.](https://tex.z-dn.net/?f=%202%2C4%20-dinitrotoluene%20%20%20%5Cfrac%7Bfuming%20HNO3-H2So4%7D%7B90-100%20C%7D%20-%3E%20%202%2C4%2C6-trinitrotoluene.%20)
So over all reaction uses three reagents in order :
![Benzene \frac{AlCl3}{CH3Cl} -> Toluene \frac{HNO3-H2So4}{room temp} -> 2,4-dinitrotoluene \frac{Fuming HNO3 -H2SO4}{Heating at 90-100 C} -> 2,4,6-trinitrotoluene .](https://tex.z-dn.net/?f=%20Benzene%20%20%5Cfrac%7BAlCl3%7D%7BCH3Cl%7D%20%20-%3E%20Toluene%20%20%5Cfrac%7BHNO3-H2So4%7D%7Broom%20temp%7D%20%20-%3E%202%2C4-dinitrotoluene%20%20%5Cfrac%7BFuming%20HNO3%20-H2SO4%7D%7BHeating%20at%2090-100%20C%7D%20%20-%3E%202%2C4%2C6-trinitrotoluene%20.%20)
Answer:
2.0 mL of 10.5 M H2O2, where H2O2 has a molar mass of 34 g/mol.
Explanation:
It is most concentrated because it contains 10.5 M of Hydrogen peroxide.
Answer: 35.4 g
Explanation:
The balanced reaction is :
![2P+3Cl_2\rightarrow 2PCl_3](https://tex.z-dn.net/?f=2P%2B3Cl_2%5Crightarrow%202PCl_3)
To calculate the moles :
![\text{Moles of phosphorous}=\frac{15.5g}{31g/mol}=0.50moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20phosphorous%7D%3D%5Cfrac%7B15.5g%7D%7B31g%2Fmol%7D%3D0.50moles)
![\text{Moles of phosphorous chloride}=\frac{50.9g}{137g/mol}=0.372moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20phosphorous%20chloride%7D%3D%5Cfrac%7B50.9g%7D%7B137g%2Fmol%7D%3D0.372moles)
According to stoichiometry :
2 moles of phosphorous chloride are produced by = 3 moles of ![Cl_2](https://tex.z-dn.net/?f=Cl_2)
Thus 0.37 moles of phosphorous chloride are produced by=
of ![Cl_2](https://tex.z-dn.net/?f=Cl_2)
Mass of ![Cl_2=moles\times {\text {Molar mass}}=0.558moles\times 71g/mol=35.4g](https://tex.z-dn.net/?f=Cl_2%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.558moles%5Ctimes%2071g%2Fmol%3D35.4g)
Thus 35.4 g of chlorine reacted with the phosphorus
To determine how much water can form would depend on what the acetic acid is reacting with. However, a very common reaction of acetic acid that leads to the formation of water is the condensation that occurs during the esterification process. We can react acetic acid with an alcohol, such as methanol, and form the ester and water.
CH₃CO₂H + CH₃OH → CH₃CO₂CH₃ + H₂O
The mole ratio of water to acetic acid is 1:1. We can now use the mass of acetic acid to find the moles, which we can convert to moles of water:
5.10 g CH₃CO₂H / 60 g/mol = 0.085 mol CH₃CO₂H x 1 mol H₂O/1 mol CH₃CO₂H = 0.085 mol H₂O
0.085 mol H₂O x 18 g/mol = 1.53 g H₂O
In the reaction provided, assuming the reaction goes to completion, 5.10 g of acetic acid can form 1.53 g of water.