Question

Consider the balanced chemical equation below:

2 NaNO3 -> 2NaNO2 + O2
How many grams of O2(g) could be produced from the reaction of 25.0 grams of NaNO3?
A. Between 6 and 8
B. Between 0 and 2
C. Between 2 and 4
D. More than 8
E. Between 4 and 6

Answer 1

By using stoichiometry and molar mass calculations, 25.0 grams of NaNO3 can produce approximately 4.704 grams of O2, placing the answer in the range of between 4 and 6 grams.

Step-by-step explanation:

Stoichiometry and Molar Mass Calculation

To determine how many grams of les of NaNO3 produce 1 mole of O2, so 0.294 mol of NaNO3 would produce 0.294 / 2 = 0.147 mol of O2.O2(g) could be produced from the reaction of 25.0 grams of NaNO3, we must first use stoichiometry and molar mass concepts. The balanced equation given is 2 NaNO3 -> 2 NaNO2 + O2, which indicates that 2 moles of NaNO3 will produce 1 mole of O2.

First, calculate the molar mass of NaNO3 (85.0 g/mol). Then, using the mass of NaNO3 provided (25.0 g), we can find the number of moles present: 25.0 g NaNO3 * (1 mol NaNO3 / 85.0 g NaNO3) = 0.294 mol NaNO3. According to the balanced chemical equation, 2 mo

Now, using the molar mass of O2 (32.0 g/mol), we can calculate the mass of O2 produced: 0.147 mol * 32.0 g/mol = 4.704 g of O2. Thus, the mass of O2 produced falls between 4 and 6 grams.