how many grams of potassium chlorate must be used to produce 30.0g of oxygen?
Here is a worked example using KClO3 and oxygen. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
115g
To determine the amount of potassium chlorate needed to produce 30.0g of oxygen, we need to know the balanced chemical equation for the decomposition of potassium chlorate (KClO3) to produce oxygen (O2).
The balanced chemical equation is:
2 KClO3 -> 2 KCl + 3 O2
From the equation, we can see that 2 moles of potassium chlorate decomposes to produce 3 moles of oxygen.
Now, we need to calculate the molar mass of oxygen (O2) to convert the given mass (30.0g) into moles.
The molar mass of O2 is given by:
Molar mass = atomic mass of O * 2
= 16.00 g/mol * 2
= 32.00 g/mol
Next, we can use the molar ratio from the balanced chemical equation to calculate the moles of potassium chlorate needed.
Moles of KClO3 = (moles of O2) * (moles of KClO3 / moles of O2)
= (30.0g / 32.00 g/mol) * (2 moles KClO3 / 3 moles O2)
Calculating this, we get:
Moles of KClO3 = (30.0g / 32.00 g/mol) * (2 / 3)
≈ 0.9375 mol
Finally, we can convert the moles of potassium chlorate into grams using its molar mass.
Grams of KClO3 = (moles of KClO3) * (molar mass of KClO3)
= 0.9375 mol * (122.55 g/mol)
≈ 114.85 g
Therefore, approximately 114.85 grams of potassium chlorate must be used to produce 30.0g of oxygen.
To determine the number of grams of potassium chlorate required to produce 30.0g of oxygen, we need to use the balanced chemical equation for the decomposition of potassium chlorate (KClO3):
2KClO3(s) -> 2KCl(s) + 3O2(g)
From the equation, we can see that every 2 moles of KClO3 produces 3 moles of O2 (oxygen). We can use this information to set up a proportion to find the number of moles of KClO3 required.
First, calculate the molar mass of oxygen (O2). The molar mass of O2 is 2 * 16.00g/mol = 32.00 g/mol.
Next, set up the proportion:
(30.0g O2) / (32.00g O2/mol O2) = (x g KClO3) / (molar mass KClO3)
Now, we need to find the molar mass of KClO3.
Potassium (K) has a molar mass of 39.10g/mol,
Chlorine (Cl) has a molar mass of 35.45g/mol, and
Oxygen (O) has a molar mass of 16.00g/mol.
Molar mass KClO3 = (39.10g K/mol K) + (35.45g Cl/mol Cl) + (3 * 16.00g O/mol O) = 122.55g/mol
Substitute the molar mass of KClO3 into the proportion:
(30.0g O2) / (32.00g O2/mol O2) = (x g KClO3) / (122.55g KClO3/mol KClO3)
Now, solve for x:
x = [(30.0g O2) * (122.55g KClO3/mol KClO3)] / (32.00g O2/mol O2)
x ≈ 114.34g KClO3
Therefore, approximately 114.34 grams of potassium chlorate must be used to produce 30.0 grams of oxygen.