10g of hyrogen react with excess of oxygen gas according to the equation:2H2+O2---->2H2O
what is the reacted volume of oxygen gas at STP and the mass of the produced water vapour in this reaction ?

Answer:

Volume of O₂ = 56 dm³

mass of water vapors (H₂O) = 90 g

Explanation:

Data Given:

mass of Oxygen = 10 g

Volume of Oxygen = ?

mass of the water vapor = ?

Reaction Given:

                 2H₂+O₂---->2H₂O

Solution:

First we have to look at the reaction for the information required

                 2H₂   +  O₂  -------> 2H₂O

               2 mol    1mol           2 mol

now convert moles to grams

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of O₂ = 2(16) = 32 g/mol

molar mass of H₂0 = 2(1) + 16 = 18 g/mol

So the masses will be

                      2H₂          +              O₂        ------->      2H₂O

                2 mol (2 g/mol)      1mol (32 g/mol)         2 mol (18 g/mol)

                      4 g                            32 g                           36 g

So now we know that

4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.

By using above information

First we find the volume of Oxygen:

For this first we find mass and then moles of Oxygen

As we know

if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen

Apply unity formula

                        4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂

                         10 g of hydrogen H₂ ≅ X g of Oxygen O₂    

by doing Cross multiplication

                         g of Oxygen O₂   = 32 g x 10 g / 4 g

                         g of Oxygen O₂   = 80 g

So,                  

mass of oxygen = 80 g

now find moles of oxygen

formula used:

            no. of moles = mass in grams/ molar mass . . . . . . (1)

Put values in above equation 1

            no. of moles = 80 g / 32 g/mol

             no. of moles = 2.5

Now to find volume of oxygen

Formula used

 Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)

Put values in equation 2

             Volume of O₂ = 2.5 moles x 22.4 dm³/mol

            Volume of O₂ = 56 dm³

______________________

Now to find mass of water vapors

As we now

if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen

Apply unity formula

                        4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)

                         10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)  

by doing Cross multiplication

                         g of water vapors (H₂O) = 36 g x 10 g / 4 g

                         g of water vapors (H₂O)   = 90 g

So,                  

mass of water vapors (H₂O) = 90 g