Solving Al + BaNO3 Reactions for N2 Volume

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  • Thread starter Thread starter Priyadarshini
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Discussion Overview

The discussion revolves around the reaction between aluminium powder and anhydrous barium nitrate, specifically focusing on calculating the volume of nitrogen gas produced when a given mass of barium nitrate reacts with excess aluminium. The scope includes homework-related problem-solving and chemical reaction equations.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant presents a balanced chemical equation for the reaction and attempts to calculate the volume of nitrogen produced from a specific mass of barium nitrate.
  • The participant calculates the molar mass of barium nitrate and the moles of nitrogen gas produced based on the stoichiometry of the reaction.
  • Another participant corrects the chemical formula for barium nitrate, indicating it should be ##\text{Ba(NO}_3)_2##.
  • A subsequent post acknowledges the correction regarding the chemical formula.
  • A different balanced equation is proposed by another participant, suggesting a different stoichiometric relationship between the reactants and products.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct balanced equation for the reaction, and there is disagreement regarding the calculations based on the initial formula presented.

Contextual Notes

The initial calculations depend on the correct interpretation of the chemical formula for barium nitrate, which is contested. The stoichiometric ratios in the proposed equations also differ, leading to potential variations in the calculated volume of nitrogen gas.

Who May Find This Useful

Students studying chemical reactions, particularly those interested in stoichiometry and gas volume calculations in chemistry.

Priyadarshini
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Homework Statement


The reaction between aluminium powder and anhydrous barium nitrate is used as the propellant in some fireworks. The metal oxides and nitrogen are the only products. Which volume of nitrogen, measured under room conditions, is produced when 0.783 g of anhydrous barium nitrate reacts with an excess of aluminium?

Homework Equations


8Al + 6BaNO3 ---> 4Al2O3 + 6BaO + 3N2

The Attempt at a Solution


Mr of BaNO3= 137.3 + 14+ (16x 3)
=199.3
Moles in 0.783g = 0.783/199.3
6 moles of barium nitrate gives 3 moles of nitrogen gas
so 2 moles give 1 mole of N2 gas
so moles of N2 gas: 0.783/199.3*2
Volume= moles * 24000
= 47.145 cm^3
But my answer is wrong.
 
Physics news on Phys.org
The chemical formula for barium nitrate is ##\text{Ba(NO}_3)_2##.
 
pizzasky said:
The chemical formula for barium nitrate is ##\text{Ba(NO}_3)_2##.
Ohhh, thank you!
 
10Al + 3Ba(NO3)2 ---> 5Al2O3 + 3 BaO + 3N2
 

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