Question

The equation for the action of heat on calcium nitrate is:

2Ca(NO3)2 2CaO + 4 NO2 + O2

(i) How many moles of NO2 are produced when 1 mole of Ca(NO3)2 decomposes?

(ii) What volume of O2 at S.T.P. will be produced on heating 65.6 g of Ca (NO3)2?

(iii) Find out the mass of CaO formed when 65.6 g of Ca(NO3)2 is heated.

(iv) Find out the mass of Ca(NO3)2, required to produce 5 moles of gaseous

products.

(v) Find out the mass of Ca(NO3)2 required to produce 44. 8 L of NO2 at S.T.P.

(Relative molecular mass of Ca(NO3)2 = 164 and of CaO = 56)​

Answer 1

Explanation:

in (I) left side is Ca(NO3)2 and right side is NO2 gas

Answer 2

The equation for the action of heat on calcium nitrate is:

2Ca(NO3)2 2CaO + 4 NO2 + O2

(i) as per given, 2 moles of Ca(NO3)2 gives 4 moles of NO2

 Therefore 1 mole of Ca(NO3)2 gives 4/2 = 2 moles of NO2

(ii) Moleculare weight of Ca(NO3)2 =164 g/mol

  Therefore 65.6 gm Ca(NO3)2 = (22.4 × 65.62) / (2 × 164)

 2 × 164 gm of Ca(NO3)2 = 22.4 liter of O2

  = 4.48 liter of O2.

(iii) 2Ca(NO3)2 --------> 2CaO + 4NO2 + O2

 (2 × 164)                   (2 × 56) (4 × 46) (2 × 16)

 (2 × 164) = 328g of Ca(NO3)2 forms ((2 × 56) = 112g of CaO

 65.6g of Ca(NO3)2 produces how many grams of CaO?

  = 112/328 x 656/10

  = 22.4 g of CaO

(iv) 5 moles of gaseous product is formed by 2 moles of Ca(NO3)2

 1 mole of Ca(NO3)2 weighs = 164g

 Then, 2 moles of Ca(NO3)2 weighs = 2 × 164 = 328 g

(v) According to the given equation

 4 × 22.4 liters of NO2 will be formed by 2 × 164g of calcium nitrate

 44.8 liters of NO2 will be formed by how many grams of calcium nitrate?

 = 2 × 164 / 4 × 22.4 × 44.8

 = 164 g of calcium nitrate