Chemistry, asked by ppsingh8434, 10 months ago

8.6 g of a metal oxide is reduced with hydrogen gas to yield 1.8g of water. The percentage (by mass)of hydrogen in metal hydride will be....

Answers

Answered by aryan1742
2

2.77%(Ans)

Explanation:

  • M2O+2H2-->2MH+H20
  • Let Mass of metal be x
  • Moles of metal oxide =8.6/(2x+16)
  • Moles of H20=1.8/18=0.1

According to stoichiometry,

Moles of metal oxide=Moles of H20

Therefore,

8.6/(2x+16)=0.1

x=35

Percentage of H in Metal hydride=(1×1×100)/(35+1)

=2.77% or 2.8%

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