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Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force.

NH4NO3(s) → N2(g) + O2(g) + H2O(g)

Calculate the mass of each product gas if 1.29 g of ammonium nitrate reacts.

N2
O2
H2O

Respuesta :

Eduard22sly

1. The mass of N₂ produced is 0.45 g

2. The mass of O₂ produced is 0.26 g

3. The mass of H₂O produced is 0.58 g

Balanced equation

4NH₄NO₃(g) --> 4N₂(g) + 2O₂(g) + 8H₂O(g)

Molar mass of NH₄NO₃ = 80 g/mol

Mass of NH₄NO₃ from the balanced equation = 4 × 80 = 320 g

Molar mass of N₂ = 28 g/mol

Mass of N₂ from the balanced equation = 4 × 28 = 112 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of H₂O = 18 g/mol

Mass of H₂O from the balanced equation = 8 × 18 = 144 g

SUMMARY

From the balanced equation above,

320 g of NH₄NO₃ decomposed to produce 112 g of N₂, 64 g of O₂ and 144 g of H₂O

1. How to determine the mass of N₂ produced

From the balanced equation above,

320 g of NH₄NO₃ decomposed to produce 112 g of N₂

Therefore,

1.29 g of NH₄NO₃ will decompose to produce = (1.29 × 112) / 320 = 0.45 g of N₂

2. How to determine the mass of O₂ produced

From the balanced equation above,

320 g of NH₄NO₃ decomposed to produce 64 g of O₂

Therefore,

1.29 g of NH₄NO₃ will decompose to produce = (1.29 × 64) / 320 = 0.26 g of O₂

3. How to determine the mass of H₂O produced

From the balanced equation above,

320 g of NH₄NO₃ decomposed to produce 144 g of H₂O

Therefore,

1.29 g of NH₄NO₃ will decompose to produce = (1.29 × 144) / 320 = 0.58 g of H₂O

Learn more about stoichiometry:

https://brainly.com/question/14735801

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