In a decomposition reaction 2KCLO3 -> 2KCL + 3O2, what volume of O2 gas at STP would be produced from 3.66g of KCLO3

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To find the volume of O2 gas produced, you need to use stoichiometry and the ideal gas law.

First, convert the mass of KCLO3 to moles:

Molar mass of KCLO3 = 39.1 g/mol (K) + 35.45 g/mol (Cl) + 3 * 16.00 g/mol (O) = 122.55 g/mol

moles of KCLO3 = mass of KCLO3 / molar mass of KCLO3
= 3.66 g / 122.55 g/mol
= 0.0299 mol

According to the balanced chemical equation, 2 moles of KCLO3 produce 3 moles of O2. Therefore, moles of O2 produced = (0.0299 mol KCLO3) * (3 mol O2 / 2 mol KCLO3) = 0.0449 mol O2

Now, we can use the ideal gas law to find the volume of O2 at STP (Standard Temperature and Pressure).

The ideal gas law is given by: PV = nRT

where:
P is the pressure (STP is 1 atm)
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (STP is 273.15 K)

Rearranging the equation to solve for V gives us: V = nRT / P

Plugging in the values:

V = (0.0449 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / 1 atm

Thus,
V = 1.090 L

Therefore, 3.66 g of KCLO3 will produce 1.090 L of O2 gas at STP.

To find the volume of O2 gas produced from 3.66g of KCLO3 at STP (Standard Temperature and Pressure), we can use the following steps:

Step 1: Determine the molar mass of KCLO3.
The molar mass of KCLO3 can be calculated by summing the atomic masses of its constituent elements:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol (there are three oxygen atoms in KCLO3)
Molar mass of KCLO3 = (39.10 g/mol) + (35.45 g/mol) + (3 * 16.00 g/mol) = 122.55 g/mol

Step 2: Convert the given mass of KCLO3 to moles.
Using the molar mass calculated in step 1, we can determine the number of moles of KCLO3:
Number of moles = given mass / molar mass
Number of moles = 3.66 g / 122.55 g/mol = 0.0299 mol (rounded to four decimal places)

Step 3: Apply stoichiometry to determine the moles of O2 gas produced.
From the balanced equation, we can see that the molar ratio between KCLO3 and O2 is 2:3. This means that for every 2 moles of KCLO3, 3 moles of O2 are produced.
Using this ratio, we can calculate the number of moles of O2 gas produced:
Moles of O2 = moles of KCLO3 * (3 moles of O2 / 2 moles of KCLO3)
Moles of O2 = 0.0299 mol * (3/2) = 0.0448 mol (rounded to four decimal places)

Step 4: Apply the ideal gas law to determine the volume of O2 gas at STP.
The ideal gas law relates the volume of a gas to its number of moles:
PV = nRT
P: pressure (at STP, P = 1 atm)
V: volume (to be determined)
n: number of moles of gas (0.0448 mol)
R: ideal gas constant (0.0821 L·atm/(mol·K))
T: temperature in Kelvin (at STP, T = 273 K)
Rearranging the equation to solve for V:
V = (nRT) / P
V = (0.0448 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm

Calculating the above equation will give us the volume of O2 gas at STP.

http://www.jiskha.com/science/chemistry/stoichiometry.html

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