Final answer:
To form 26.4 grams of water from the decomposition of ammonium nitrite, 58.60 grams of ammonium nitrite are required based on stoichiometric calculations.
Step-by-step explanation:
To solve the problem of how many grams of ammonium nitrite are needed to form 26.4 grams of water, we must use stoichiometry based on the chemical equation for the decomposition of ammonium nitrite:
NH₄NO₂ (s) → N₂O(g) + 2H₂O (1)
Firstly, we identify the molar masses:
- 1 mol NH₄NO₂ = 80.06 g/mol
- 1 mol H₂O = 18.02 g/mol
Because the ratio of ammonium nitrite to water in moles is 1:2, we can use this ratio to find the number of moles of water produced and work our way back to find the mass of ammonium nitrite required.
Using the molar mass of water (18.02 g/mol), we convert the given mass of water to moles:
26.4 g H₂O x (1 mol H₂O / 18.02 g H₂O) = 1.464 moles H₂O
Since for every 2 moles of water, 1 mole of ammonium nitrite is decomposed, we have:
1.464 moles H₂O x (1 mol NH₄NO₂ / 2 moles H₂O) = 0.732 moles NH₄NO₂
Finally, we convert moles of ammonium nitrite to grams:
0.732 moles NH₄NO₂ x (80.06 g NH₄NO₂ / 1 mol NH₄NO₂) = 58.60 grams NH₄NO₂
Therefore, 58.60 grams of ammonium nitrite are needed to form 26.4 grams of water, respecting the law of conservation of mass.