Question

For a reaction:

Answer 1

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

$\\ \sf\longmapsto pV=nRT$

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

$\\ \sf\longmapsto 1.5V=448.2$

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

$\\ \sf\longmapsto V=298.8mL$