Question

What total volume, in litre at 600°C and 1 atm, could be formed by the decomposition of 16 g of NH4NO3 ?
Reaction : 2 NH4NO3  2N2 + O2 + 4H2O(g).
ANS PLZ!!!!!!

Answer 1

Answer:

Ammonium Nitrate is decomposed as,

NH4NO3 -> N2O + 2H2O,

According to stoichiometric coeffiecients, 1 mole of NH4NO3 dissociates to form 1 mole of nitrous oxide and 2 moles of water, but in the question we have 16g of NH4NO3,

So the number of moles of NH4NO3 = 16/mass of 1 mole of NH4NO3 = 16/80 = 0.2 mole,

Use the ideal gas equation PV = nRT,

V = (0.2)(0.082)(873)/1 = 14.3172 L

Explanation:

Hope It'll help u!

Answer 2

Answer:

The molar mass of ammonium nitrate is 80g/mol.

2 moles of ammonium nitrate on decomposition gives 7 moles of product.

Hence, 16g of ammonium nitrate on decomposition will give

80×2

16×7

=0.7 moles of product.

This corresponds to 0.7×22.4=15.68L at STP.

At 627

∘

C, the total volume is equal to V

2

=

T

1

T

2

×V

1

=

273

900

×15.68=51.723L .

no I'm from Delhi...