Question

What mass of potassium nitrate is needed to generate 199.0 L of gas, composed of 141.0 l of N₂ and 58.0 l of O₂ at 0.920 atm and 299 k, using these two reactions?

Answer 1

To generate 199.0 L of gas composed of N₂ and O₂, a mass of 608.8 g of potassium nitrate is needed.

Step-by-step explanation:

To determine the mass of potassium nitrate needed to generate a certain volume of gas, we need to use stoichiometry. The balanced equation for the reaction is:

4 KNO3(s) ⟶ 4 KNO2(s) + 2 O2(g)

From the balanced equation, we can see that 4 moles of KNO3 produce 2 moles of O2. So, we need to calculate the number of moles of O2 first. Using the ideal gas law, we can then convert the number of moles of O2 to the volume of O2 at the given conditions. Finally, we can use the volume of O2 to calculate the number of moles of KNO3 needed.

Let's calculate:

Given:

• Volume of O2 (VO2) = 58.0 L
• Partial pressure of O2 (PO2) = 0.920 atm
• Temperature (T) = 299 K
• Ideal gas constant (R) = 0.0821 L·atm/(K·mol)

First, we calculate the number of moles of O2 using the ideal gas law:

n(O2) = (P(V)/RT) = (0.920 atm × 58.0 L)/(0.0821 L·atm/(K·mol) × 299 K) ≈ 3.01 moles

Next, we use the stoichiometry of the reaction to calculate the number of moles of KNO3:

n(KNO3) = (4/2) × n(O2) = 4/2 × 3.01 moles = 6.02 moles

Finally, we convert the number of moles of KNO3 to its mass:

Mass of KNO3 = n(KNO3) × molar mass of KNO3 = 6.02 moles × 101.1 g/mol = 608.8 g