Question

For a reaction:
Ca(OH)₂ + 2NH₂Cl → CaCl₂ + 2NH₂ + 2H₂O

The reaction is carried out by mixing 7 g of pure (Ca(OH)₂ and 7 g of pure NHCI
(a) Find the limiting reactant.
(b) Calculate the mole of unreacted reactant left over.
(c) How many grams of CaCl, are formed? (d) What volume of NH, gas are produced at 27°C and 1.5 atmospheric pressure?​

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

$\\ \sf\longmapsto pV=nRT$

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

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

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

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