Hydrogen-filled balloon was ignited and 1.50 g of hydrogen reacted with 12.0 g of oxygen. how many grams of water vapor formed? (assume that water vapor is the only product.)

             13.39 g of H₂O

Solution:

The Balance Chemical Reaction is as follow,

                                          2 H₂  +   O₂    →    2  H₂O

Step 1: Find out the limiting reagent as;

According to Equation ,

            4.032 g (2 mole) H₂ reacts with  =  32 g (1 moles) of O₂

So,

                  1.5 g of H₂ will react with  =  X g of O₂

Solving for X,

                      X  =  (1.5 g × 32 g) ÷ 4.032 g

                      X  =  11.90 g of O₂

It means for total oxidation of Hydrogen we require 11.90 g of O₂, but we are provided with 12 g of O₂. Therefore, H₂ is the limiting reagent and will control the yield.

Step 2: Calculate Amount of H₂O produced as;

According to Equation ,

              4.032 g (2 mole) H₂ produces  =  36 g (2 moles) of H₂O

So,

                       1.50 g of H₂ will produce  =  X g of H₂O

Solving for X,

                      X =  (1.50 g × 36 g) ÷ 4.032 g

                      X  =  13.39 g of H₂O

And this answer is correct.

just did the assignment and the answer was #1 +2 n

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.