Find the volume of oxygen produced by 1mole of potassium trioxochlorate(v) at stp in the following reaction
2KCLO3➡️2KCL +3O2
According to the balanced chemical equation, 2 moles of potassium chlorate (KCLO3) react to produce 3 moles of oxygen gas (O2).
Therefore, for every 2 moles of potassium chlorate, 3 moles of oxygen gas are produced.
Since we want to find the volume of oxygen gas produced by 1 mole of potassium chlorate, we need to calculate the molar volume of oxygen gas at STP (Standard Temperature and Pressure).
The molar volume of a gas at STP is approximately 22.4 liters/mol.
Therefore, if 2 moles of potassium chlorate produce 3 moles of oxygen gas, then 1 mole of potassium chlorate will produce (3/2) moles of oxygen gas.
To find the volume of oxygen gas produced by 1 mole of potassium chlorate, we multiply the number of moles by the molar volume:
Volume = (3/2) moles * 22.4 liters/mol = 33.6 liters
Thus, the volume of oxygen gas produced by 1 mole of potassium trioxochlorate(V) at STP is 33.6 liters.
To find the volume of oxygen produced by 1 mole of potassium trioxochlorate(V) (KClO3) at STP (Standard Temperature and Pressure), you need to use the molar volume of gases at STP.
The molar volume of gases at STP is 22.4 liters/mole.
From the balanced equation:
2KClO3 ➡️ 2KCl + 3O2
You can see that for every 2 moles of KClO3, 3 moles of O2 are produced.
So, if 2 moles of KClO3 produce 3 moles of O2, then 1 mole of KClO3 will produce (3/2) moles of O2.
Now, you can calculate the volume of O2 produced by 1 mole of KClO3 at STP:
Volume of O2 = (Number of moles of O2) x (Molar volume of gases at STP)
= (3/2) moles x 22.4 liters/mole
= 33.6 liters of O2
Therefore, the volume of oxygen produced by 1 mole of potassium trioxochlorate(V) at STP in the given reaction is 33.6 liters.
To find the volume of oxygen produced by 1 mole of potassium trioxochlorate(V) at STP (Standard Temperature and Pressure), we can use the ideal gas law equation, PV = nRT.
In this reaction, we can see that 2 moles of potassium trioxochlorate(V) (KCLO3) produces 3 moles of oxygen (O2).
Since we're given 1 mole of KCLO3, we need to find the volume of oxygen produced.
Step 1: Convert moles of oxygen to volume using the ideal gas law equation.
PV = nRT
We'll assume the pressure (P) is 1 atm and the temperature (T) is 273 K (STP).
Step 2: Determine the gas constant (R). The ideal gas constant is typically given as R = 0.0821 L·atm/(mol·K).
Step 3: Calculate the volume(V).
V = nRT/P
In this case, n = 3 moles (since 1 mole of KCLO3 produces 3 moles of O2), P = 1 atm, R = 0.0821 L·atm/(mol·K), and T = 273 K.
V = (3 mol)(0.0821 L·atm/(mol·K))(273 K) / (1 atm)
V = 61.53 L
Therefore, the volume of oxygen produced by 1 mole of potassium trioxochlorate(V) at STP is 61.53 liters.