How many mL of O2 will form from 55.2g KClO3?
Here is a worked example. Just follow the steps.
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How many mL of O2 will form from 55.2g KClO3?
To determine the number of mL of O2 that will form from 55.2g of KClO3, we need to use the molar mass of KClO3 and the balanced chemical equation.
The molar mass of KClO3 is:
K - 39.10 g/mol
Cl - 35.45 g/mol
O - 16.00 g/mol
Molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + (3 * 16.00 g/mol)
= 122.55 g/mol
Next, we need to use the balanced chemical equation for the decomposition of KClO3:
2 KClO3 -> 2 KCl + 3 O2
From the equation, we can see that for every 2 moles of KClO3, we produce 3 moles of O2.
Now, we can set up a conversion using the molar mass and stoichiometry:
55.2 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) * (3 mol O2 / 2 mol KClO3) * (22.4 L O2 / 1 mol O2) * (1000 mL / 1 L) = X mL O2
Calculating this expression will give us the number of mL of O2 formed. Let's solve:
55.2 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) * (3 mol O2 / 2 mol KClO3) * (22.4 L O2 / 1 mol O2) * (1000 mL / 1 L) = 2500 mL O2
Therefore, from 55.2g of KClO3, approximately 2500 mL of O2 will form.
To find out how many milliliters (mL) of O2 will form from 55.2g of KClO3 (potassium chlorate), we will need to use the molar ratio between KClO3 and O2.
First, let's calculate the number of moles of KClO3 using its molar mass. The molar mass of KClO3 is the sum of the atomic masses of potassium (K), chlorine (Cl), and three oxygen atoms (O).
Molar mass of KClO3:
K = 39.10 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol (three O in KClO3)
Molar mass of KClO3 = 39.10 + 35.45 + (16.00 * 3) = 122.55 g/mol
Now, we can calculate the number of moles of KClO3 using its mass and molar mass:
moles of KClO3 = mass / molar mass
moles of KClO3 = 55.2 g / 122.55 g/mol ≈ 0.4507 mol (rounded to four decimal places)
Next, we need to use the balanced chemical equation for the reaction between KClO3 and O2 to determine the mole ratio between KClO3 and O2. The balanced equation is:
2 KClO3 → 2 KCl + 3 O2
From the equation, we can see that for every 2 moles of KClO3, 3 moles of O2 are produced.
Now, we can use the mole ratio to calculate the moles of O2 formed:
moles of O2 = moles of KClO3 × (3 moles of O2 / 2 moles of KClO3)
moles of O2 = 0.4507 mol × (3/2) ≈ 0.6761 mol (rounded to four decimal places)
Finally, to find the volume of O2 in milliliters (mL), we need to apply the ideal gas law, which states that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). To convert to milliliters, we can use the conversion:
1 L = 1000 mL
Therefore, the volume of O2 in mL can be calculated as:
volume of O2 = moles of O2 × 22.4 L/mol × 1000 mL/L
volume of O2 = 0.6761 mol × 22.4 L/mol × 1000 mL/L ≈ 15,135.04 mL (rounded to two decimal places)
So, approximately 15,135.04 mL of O2 will form from 55.2g of KClO3.