Answer:
A 1 liter volumetric flask should be used.
Explanation:
First we <u>convert 166.00 g of KI into moles</u>, using its <em>molar mass</em>:
Molar mass of KI = Molar mass of K + Molar mass of I = 166 g/mol
- 166.00 g ÷ 166 g/mol = 1 mol KI
 
Then we <u>calculate the required volume</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
 
Liters = moles / molarity
 
        
             
        
        
        
For the first one the answer is B. and the second one is D.
        
                    
             
        
        
        
The balanced chemical equation of the reaction described above is,
                            C2H6O + O2 --> H2O  + C2H4O2
If we have 3.84 g of oxygen, we divide by its molar mass.
 
                               n = (3.54 g Oxygen gas) x (1 mole O2/ 32 g O2)
                                 n = 0.11 moles O2
Using ratio and proportion,
               number of moles of ethanol = (0.11 moles O2) x (1 mole C2H6)
                                                  = 0.11 moles C2H6
Then, we multiply the calculated value to its molar mass, 46 grams /mol.
                    mass of ethanol = (0.11 mol) x (46 grams / mol)
                                                = <em>5.06 grams</em>
        
             
        
        
        
Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors. 
<span>                               2 Al(OH)</span>₃    →    Al₂O₃  +  3 H₂O
According to equation at STP,
       67.2 L (3 moles) of H₂O is produced by  =  78 g of Al(OH)₃
So,
                65.0 L of H₂O will be produced by  =  X g of Al(OH)₃
Solving for X,
                                 X  =  (65.0 L × 78 g) ÷ 67.2 L
                                 X  =  
75.44 g of Al(OH)₂Result:           75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry