Final answer:
To determine how many moles of nitrogen gas are formed from 30.5 grams of ammonium nitrite, we calculate the molar mass of ammonium nitrite, divide the given mass by this molar mass to find the number of moles, and then apply the stoichiometry of the decomposition reaction. The result is 0.476 moles of nitrogen gas.
Step-by-step explanation:
The subject of the question revolves around stoichiometry, a concept in chemistry that involves the calculation of reactants and products in chemical reactions. The specific stoichiometric calculation deals with ammonium nitrite decomposing into nitrogen gas and water. To determine the amount of nitrogen gas formed from the reaction, we must use the reaction's molar mass and stoichiometry.
First, calculate the molar mass of ammonium nitrite (NH4NO2). Ammonium nitrite consists of 2 nitrogen atoms, 4 hydrogen atoms, and 2 oxygen atoms. Using the periodic table, the atomic weights are approximately 14.01 g/mol for nitrogen, 1.01 g/mol for hydrogen, and 16.00 g/mol for oxygen. This gives us a molar mass of (14.01 g/mol * 2) + (1.01 g/mol * 4) + (16.00 g/mol * 2) = 64.06 g/mol.
Next, we need to find out how many moles of ammonium nitrite the 30.5 grams represent. This is achieved by dividing the mass of ammonium nitrite by its molar mass: 30.5 g / 64.06 g/mol = 0.476 moles of NH4NO2.
The reaction for the decomposition of ammonium nitrite is as follows:
NH4NO2(s) → N2(g) + 2H2O(g).
According to the balanced equation, one mole of ammonium nitrite produces one mole of nitrogen gas. Therefore, if we have 0.476 moles of NH4NO2, the same amount (0.476 moles) of N2 will be produced upon complete reaction.