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jeyben [28]
3 years ago
15

For a reaction:

Chemistry
1 answer:
torisob [31]3 years ago
7 0

Looking at the equation .look at the moles noted below

  • Ca(OH)_2=1mol
  • NH_2Cl=2mol.
  • CaCL_2=1mol
  • NH_2=2mol
  • H_20=2mol.

#a

There is 2 moles of NH2Cl.

  • Hence limiting reagent is Ca(OH)_2

#2

Moles at reactant=3

Moles at product=5

Moles left:-

\\ \sf\longmapsto 5-3=2mol

#d

\\ \sf\longmapsto pV=nRT

\\ \sf\longmapsto 1.5V=2(8.3)(27)

\\ \sf\longmapsto 1.5V=448.2

\\ \sf\longmapsto V=\dfrac{448.2}{1.5}

\\ \sf\longmapsto V=298.8mL

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