Final answer:
The expected number of moles of nitrogen produced from 0.0928 g of sodium nitrite is 0.0006725 moles, calculated using stoichiometry based on the molar mass of NaNO2 and the balanced chemical equation.
Step-by-step explanation:
To find the expected number of moles of nitrogen produced from 0.0928 g of sodium nitrite (NaNO2), we need to know the chemical equation for the reaction. Assuming the decomposition reaction of sodium nitrite to produce nitrogen is as follows:
2 NaNO2 → 2 NaNO3 + N2
Now, calculating the amount of nitrogen involves finding the molar mass of NaNO2 and then using stoichiometry to determine moles of nitrogen produced.
The molar mass of NaNO2 is approximately 69 grams per mole (22.99 g/mol for Na + 14.01 g/mol for N + 32.00 g/mol for O2). Using this molar mass:
0.0928 g NaNO2 × (1 mol NaNO2 / 69 g NaNO2) = 0.001345 mol NaNO2
From the balanced equation, 2 moles of NaNO2 produce 1 mole of N2. So:
0.001345 mol NaNO2 × (1 mol N2 / 2 mol NaNO2) = 0.0006725 mol N2
Thus, 0.0006725 moles of nitrogen gas can be expected to be produced under these conditions.