Answer:
Mass of water produced is 22.86 g.
Explanation:
Given data:
Mass of hydrogen = 2.56 g
Mass of oxygen = 20.32 g
Mass of water = ?
Solution:
Chemical equation:
2H₂ + O₂   →  2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 20.32 g/ 32 g/mol
Number of moles = 0.635 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 2.56 g/ 2 g/mol
Number of moles = 1.28 mol
Now we will compare the moles of water with oxygen and hydrogen.
                     O₂            :            H₂O
                      1              :             2
                   0.635        ;            2×0.635 =  1.27
                    H₂             :              H₂O
                     2              :              2
                  1.28            :           1.28 
The number of  moles of water produced by oxygen are less thus it will be limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 1.27 × 18 g/mol
Mass = 22.86 g
   
 
        
                    
             
        
        
        
Answer:
4 moles of water
Explanation:
this is a combustion reaction, so the balanced equation is: 2C2H6 + 7O2 → 4CO2 + 6H2O.
the molar mass of C2H6 is 30.07g, so 40.0 g of C2H6 is 1.33 moles of C2H6. 
mole ratio of H2O to C2H6 is 6/2, or 3.
1.33 moles C2H6 * 3 moles H2O/1 mole C2H6 = 4 moles H2O 
 
        
             
        
        
        
Answer:
A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity. 
Explanation:
We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.
In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.
 
        
             
        
        
        
The correct answer is Be+
That is because it lost a single electron but still has the same number of protons, and thus the effective charge attracting each electron is greater, which in turn makes the radius even smaller