Question

10 g of hydrogen and 64g of oxygen were filled in a steel vessel and exploded . volume of gaseous product after reaction

Answer 1

Answer: hello buddy friend the answer is :-

The balanced equation for this reaction is as follows: 

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

Find the moles of the reacting gases.

10g of hydrogen

moles = mass/molar mass

         = 10g / 2

         =  5 moles

64g of oxygen

mass/molar mass

moles = 64/32

 = 2 moles

The mole reaction for the equation above is H₂ : O₂  is 2 :1

Therefore 5 moles of H₂ will require 2.5 moles of oxygen - this makes oxgen the limiting reagent.

We therefore use the moles of oxygen to find those of H₂O:

mole ratio  O₂ : H₂O   is 1 : 2

That means 2 moles of oxygen will produce 4 moles of H₂O

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Calculate volume of 4 moles of H₂O gas.

Ideal gas law states that 1 mole of an ideal gas occupies exactly 22.4 liters by volume.

If 1 mole = 22.4 liters

Then 4 moles = 22.4 × 4/1

                      = 89.6 liters

The volume of H₂O produced is 89.6 liters.

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Answer 2

answer

4 gram H2 requires 36 g of O2

then 10 g h2 requires 90 g of oxygen.so oxygen is limiting reagent