what volume of dry oxygen gas measured at stp will be produced from the decomposition of 3.50 of potassium trioxochlorate (v)

To calculate the volume of dry oxygen gas produced from the decomposition of potassium trioxochlorate (V), you need to use the balanced chemical equation for the reaction.

The balanced equation for the decomposition of potassium trioxochlorate (V) is:

2KClO3(s) → 2KCl(s) + 3O2(g)

According to the stoichiometry of the balanced equation, for every 2 moles of KClO3 decomposed, 3 moles of O2 gas are produced.

Now, let's calculate the molar amount of KClO3:

Given: Mass of KClO3 = 3.50 g
Molar mass of KClO3 = 122.55 g/mol

Number of moles of KClO3 = Mass / Molar mass
= 3.50 g / 122.55 g/mol
≈ 0.0285 mol

Since 2 moles of KClO3 produce 3 moles of O2 gas, we need to use the ratio to find the moles of O2 gas:

Number of moles of O2 = Number of moles of KClO3 × (3 moles O2 / 2 moles KClO3)
= 0.0285 mol × (3/2)
≈ 0.0427 mol

Now we can calculate the volume of O2 gas at STP (Standard Temperature and Pressure).

STP conditions are defined as:
Temperature (T) = 273.15 K
Pressure (P) = 1 atmosphere (atm)
Molar volume of gas at STP (V) = 22.4 L/mol

Using the ideal gas law equation: PV = nRT, we can solve for V:

V = nRT/P

V = (0.0427 mol)(0.0821 L·atm/(mol·K))(273.15 K)/(1 atm)
≈ 0.978 L

Therefore, approximately 0.978 liters of dry oxygen gas will be produced from the decomposition of 3.50 grams of potassium trioxochlorate (V) at STP.

To determine the volume of dry oxygen gas produced from the decomposition of potassium trioxochlorate (V), we need to consider the balanced chemical equation for this reaction. The decomposition reaction of potassium trioxochlorate (V) is as follows:

2KClO3(s) -> 2KCl(s) + 3O2(g)

From this equation, we can see that for every 2 moles of KClO3 decomposed, 3 moles of O2 gas are produced.

To calculate the volume of O2 gas produced, we need to use the ideal gas law equation, which states:

PV = nRT

Where:
P = Pressure (STP - standard temperature and pressure: 1 atm)
V = Volume (what we want to find)
n = Number of moles (what we need to calculate)
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (STP: 273.15 K)

We are given that the mass of KClO3 is 3.50 grams. First, we need to convert the mass of KClO3 into moles. The molar mass of KClO3 can be calculated as follows:

Molar mass of KClO3 = atomic mass of K (39.1 g/mol) + atomic mass of Cl (35.5 g/mol) + 3 x atomic mass of O (16 g/mol)
= 39.1 + 35.5 + 3 x 16
= 39.1 + 35.5 + 48
= 122.6 g/mol

Number of moles = mass / molar mass
= 3.50 g / 122.6 g/mol
= 0.0285 mol

According to the balanced equation, 2 moles of KClO3 produce 3 moles of O2 gas. Therefore, we need to calculate the number of moles of O2 gas produced using the ratio from the balanced equation:

Number of moles of O2 gas = (0.0285 mol KClO3) x (3 mol O2 / 2 mol KClO3)
= 0.04275 mol O2

Finally, we can substitute the values into the ideal gas law equation to find the volume of O2 gas:

PV = nRT
V = (nRT) / P
V = (0.04275 mol) x (0.0821 L·atm/(mol·K)) x (273.15 K) / 1 atm
V ≈ 0.92 L

Therefore, approximately 0.92 L of dry oxygen gas measured at STP will be produced from the decomposition of 3.50 g of potassium trioxochlorate (V).