potassium chlorate will decompose to produce potassium chloride and oxygen gas . if you begin with 45 kg of potassium chlorate, how many liters of oxygen gas measured at stp might you expect to produce from this reaction
Here is a worked example of this kind of problem. Just follow the steps. After you have mols O2 formed remember mols x 22.4L = volume O2 produced at STP.
To determine the number of liters of oxygen gas produced from the decomposition of potassium chlorate, we'll need to follow these steps:
Step 1: Determine the molar mass of potassium chlorate (KClO3).
K (potassium) = 39.1 g/mol
Cl (chlorine) = 35.5 g/mol
O (oxygen) = 16.0 g/mol
So, the molar mass of KClO3 = 39.1 + 35.5 + (3 * 16.0) = 122.6 g/mol
Step 2: Convert the mass of potassium chlorate (45 kg) to moles.
Given that the mass is in kilograms, we need to convert it to grams first.
45 kg = 45,000 grams
Now, we can convert grams to moles using the molar mass:
Number of moles = mass (g) / molar mass (g/mol)
Number of moles = 45,000 g / 122.6 g/mol
Step 3: Determine the stoichiometry of the reaction.
From the balanced equation for the decomposition of potassium chlorate, we find that for every 2 moles of KClO3, we produce 3 moles of O2.
Step 4: Calculate the moles of oxygen gas (O2) produced.
Number of moles of O2 = Number of moles of KClO3 x (3 moles O2 / 2 moles KClO3)
Step 5: Calculate the volume of gas at STP.
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters.
So, the volume of gas at STP can be calculated using the following equation:
Volume (liters) = Number of moles x 22.4
By substituting the value of the moles of O2 from Step 4 into the equation, we can find the volume of oxygen gas in liters.
This calculation will give us the answer you are looking for.