PLease help
how many grams of potassium chlorate decomposes to potassium chloride and 638 mL of O2 at 128°C and 752 torr?
the formula given:
2KClO3(s) 2KCl(s) + 3O2(g)
Here is a solved example of the problem you have. Only the numbers are different. You will need to find moles O2 by using PV = nRT
oops.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of grams of potassium chlorate that decomposes to form potassium chloride and 638 mL of oxygen gas (O2), you need to use stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction.
The given balanced chemical equation is:
2KClO3(s) → 2KCl(s) + 3O2(g)
First, convert the given volume of oxygen gas from milliliters (mL) to liters (L) since stoichiometry calculations require measurements in consistent units. We know that 1 L is equal to 1000 mL, so:
638 mL ÷ 1000 mL/L = 0.638 L
Next, calculate the number of moles of oxygen gas using the ideal gas law equation: PV = nRT, where P is the pressure in atmospheres (atm), V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (K).
Convert the given temperature from degrees Celsius to Kelvin:
Added 273 to 128 degrees °C = 401 K
Now, let's solve for the number of moles of O2:
P = 752 torr (convert to atm by dividing by 760 mmHg/atm)
V = 0.638 L
T = 401 K
R = 0.0821 L·atm/mol·K
P = nRT → n = PV / RT
n = (752 atm × 0.638 L) / (0.0821 L·atm/mol·K × 401 K)
n ≈ 11.87 moles (rounded to two decimal places)
According to the balanced equation, the ratio of moles of KClO3 to moles of O2 is 2:3. Therefore, you need two moles of KClO3 to produce three moles of O2.
Setting up a proportion:
2 moles KClO3 / 3 moles O2 = X moles KClO3 / 11.87 moles O2
Now, solve for X:
X = (2 moles KClO3 × 11.87 moles O2) / 3 moles O2
X ≈ 7.91 moles of KClO3 (rounded to two decimal places)
Finally, calculate the mass of KClO3 using the molar mass of KClO3 (potassium chlorate), which is 122.55 g/mol:
mass = moles × molar mass
mass = 7.91 moles × 122.55 g/mol
mass ≈ 968.53 grams (rounded to two decimal places)
Therefore, approximately 968.53 grams of potassium chlorate decomposes to form potassium chloride and 638 mL of oxygen gas at 128°C and 752 torr.