Final answer:
To determine the amount of sodium nitrate needed to produce 4.2 moles of diatomic oxygen, we must consider the stoichiometry of the reaction. Given the molar mass of sodium nitrate, the correct amount would be 714 grams, which does not match with any of the options provided.
Step-by-step explanation:
To find out how many grams of sodium nitrate are required to produce 4.2 moles of diatomic oxygen, we need to follow the stoichiometry of the reaction where sodium nitrate decomposes to yield sodium nitrite and oxygen gas. The balanced chemical equation for the decomposition of sodium nitrate to oxygen and sodium nitrite (NaNO2) is:
2 NaNO3 → 2 NaNO2 + O2
According to this equation, two moles of NaNO3 produce one mole of O2. Therefore, to produce 4.2 moles of O2, we would need:
(4.2 moles O2) × (2 moles NaNO3 / 1 mole O2) = 8.4 moles NaNO3 needed
The molar mass of NaNO3 is approximately 85 g/mol (23 g/mol for sodium, plus 14 g/mol for nitrogen, plus 3× 16 g/mol for oxygen), so the mass of NaNO3 required is:
(8.4 moles) × (85 g/mol) = 714 g of NaNO3
However, none of the options given in the question matches this result. It may be possible that the example provided or the question itself contains incorrect or incomplete information. Please verify the accuracy of the question or consult additional resources for the correct calculation.