Question

A number of reactions can be used to generate common gases on a laboratory scale. For example, nitrogen can be produced from sodium metal and potassium nitrate as indicated by the balanced equation: 10Na(s) + 2KNO₃(s) → K₂O(s) + 5Na₂o(s) +N₂(g) A common laboratory-scale reaction can also generate oxygen gas by heating potassium chlorate, as indicated by the balanced equation: 2KC10₃(s) = 2KCI(s)+30₂(g). What mass of potassium nitrate is needed to generate 193.0 L of gas, composed of 117.0 Lof N₂ and 76.0 Lof O₂ at 0.920 atm and 299 K, using these two reactions? _________g KNO³

Answer 1

To find the mass of potassium nitrate required to generate a mixture of gases at given conditions, one must use the ideal gas law to calculate moles needed and then apply stoichiometry to derive the mass from the balanced chemical equation.

Step-by-step explanation:

To calculate the mass of potassium nitrate (KNO3) needed to generate 193.0 L of gas at 0.920 atm and 299 K, we need to apply the ideal gas law (PV=nRT), where P is pressure, V is volume, n is moles of gas, R is the universal gas constant, and T is the temperature in Kelvin. First, let's calculate the moles of N2 produced using the ideal gas law.

VN2 = 117.0 L, P = 0.920 atm, T = 299 K

Moles of N2 (nN2) = PV / RT

Using R = 0.0821 L·atm/(mol·K),
nN2 = (0.920 atm × 117.0 L) / (0.0821 L·atm/(mol·K) × 299 K)

Next, determine the mass of KNO3 needed to generate the calculated moles of N2, considering the stoichiometry of the given balanced reaction. Since the balanced equation indicates that from 2 moles of KNO3, 1 mole of N2 is produced, we can find the mass of KNO3 required.

Now, to find the total mass of KNO3 needed for our final mix of gases, we'll combine the masses calculated for N2 and O2 production separately since they come from different reactions. We don't directly calculate the mass for O2 production here from potassium chlorate (KClO3), but it's assumed that you'd repeat a similar stoichiometric process for that part of the problem.

In such a way, you determine the amount of reactant needed to produce a desired amount of product through stoichiometry based on balanced chemical equations and the ideal gas law.

Answer 2

We will need 13393 grams of potassium nitrate to generate the given gas mixture.

we find the moles of each gas to use the balanced chemical equations and use the ideal gas law to convert volumes to moles.

V₁ / n₁ = V₂ / n₂

V₁ and V₂ are the volumes of N₂ and O₂, respectively (in L)

n₁ and n₂ are the moles of N₂ and O₂, respectively (in mol)

For N₂:

117.0 L / n₁ = 193.0 L / 0.5 mol

n₁ = 66.2 mol

For the O₂:

76.0 L / n₂ = 193.0 L / 0.75 mol

n₂ = 57.1 mol

66.2 mol N₂ * (2 mol KNO₃ / 1 mol N₂) = 132.4 mol KNO₃

From the balanced equation for O₂ production:

2 mol of KClO₃ will yield 3 mol O₂

57.1 mol O₂ * (2 mol KClO₃ / 3 mol O₂) * (1 mol KNO₃ / 2 mol KClO₃)

= 38.1 mol KNO₃

We then Convert moles of KNO₃ to mass:

mass of potassium nitrate= molar mass * no. of moles

132.4 mol KNO₃ * 101.1 g/mol = 13392.84 g

= 13393 g