How many grams of potassium chlorate, KClO3(s) must decompose to produce potassium chloride, KCl(s), and 1.45 grams of oxygen gas?
To determine the amount of potassium chlorate that must decompose, we need to use the balanced chemical equation for the reaction:
2KClO3(s) → 2KCl(s) + 3O2(g)
From the equation, we can see that 2 moles of KClO3 decompose to produce 3 moles of O2. Therefore, we can set up the following ratio:
2 moles KClO3 : 3 moles O2
To find the moles of O2 produced, we can use the molar mass of O2, which is approximately 32 g/mol. Thus, 1.45 grams of O2 is equal to:
1.45 g O2 × (1 mol O2 / 32 g O2) ≈ 0.0453 mol O2
Using the ratio from the balanced equation, we can then calculate the moles of KClO3 decomposed:
0.0453 mol O2 × (2 mol KClO3 / 3 mol O2) ≈ 0.0302 mol KClO3
Finally, to find the mass of KClO3 decomposed, we can convert the moles of KClO3 to grams using its molar mass, which is approximately 122.55 g/mol:
0.0302 mol KClO3 × (122.55 g KClO3 / 1 mol KClO3) ≈ 3.70 grams KClO3
Therefore, approximately 3.70 grams of potassium chlorate must decompose to produce 1.45 grams of oxygen gas.
To find out the mass of potassium chlorate that must decompose, we need to consider the balanced chemical equation for the reaction. In this case, the equation is:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
From the equation, we can see that 2 moles of potassium chlorate decompose to form 2 moles of potassium chloride and 3 moles of oxygen gas.
First, we need to find the molar mass of oxygen gas (O2) and potassium chloride (KCl):
- The molar mass of oxygen (O) is 16.00 g/mol. Since there are two oxygen atoms in O2, the molar mass of O2 is 2 * 16.00 g/mol = 32.00 g/mol.
- The molar mass of potassium (K) is 39.10 g/mol, and the molar mass of chlorine (Cl) is 35.45 g/mol. Therefore, the molar mass of potassium chloride (KCl) is 39.10 g/mol + 35.45 g/mol = 74.55 g/mol.
Now, let's calculate the number of moles of oxygen gas (O2) from the given mass of 1.45 grams:
- Number of moles of O2 = mass of O2 / molar mass of O2
= 1.45 g / 32.00 g/mol
≈ 0.0453 mol
Since 2 moles of potassium chlorate decompose to produce 3 moles of oxygen gas, we can now calculate the number of moles of potassium chlorate (KClO3) required to produce 0.0453 moles of oxygen gas:
- Number of moles of KClO3 = (0.0453 mol O2) * (2 mol KClO3 / 3 mol O2)
≈ 0.0302 mol KClO3
Finally, let's calculate the mass of potassium chlorate (KClO3) needed to produce 0.0302 moles:
- Mass of KClO3 = number of moles of KClO3 * molar mass of KClO3
= 0.0302 mol * (39.10 g/mol + 35.45 g/mol + 3 * 16.00 g/mol)
≈ 2.83 grams
Therefore, approximately 2.83 grams of potassium chlorate must decompose to produce 1.45 grams of oxygen gas and potassium chloride.