Write the balanced equation given that Mg(ClO3)2 decomposes into chloride and oxygen:
2Mg(ClO3)2 --> 2MgCl2 + 3O2
So if 0.302 grams of O2 is lost from 1.890 grams of a mixture of Mg(ClO3)2 and an inert material, what is the percentage of Mg(ClO3)2 in the mixture?
moles O2 lost = 0.302g/molar mass O2 = about 0.009 (that's an estimate).
Convert mols O2 to mols Mg(ClO3)2 with the coefficients.
0.009 x (2 mol Mg(ClO3)2/3 mols O2) = 0.009 x 2/3 = about 0.006
g Mg(ClO3)2 = mols x molar mass
Then % Mg(ClO3)2 = [g Mg(ClO3)2/mass sample)*100 = ?
To find the percentage of Mg(ClO3)2 in the mixture, we need to first calculate the initial amount of Mg(ClO3)2 in the mixture and then compare it to the total mass of the mixture.
Here's how you can do it step by step:
1. Calculate the initial mass of Mg(ClO3)2 in the mixture:
- We know that 0.302 grams of O2 was lost from the mixture.
- From the balanced equation, we can see that for every 3 moles of O2 produced, we need 2 moles of Mg(ClO3)2.
- Calculate the molar mass of O2: 32 grams/mol.
- Determine the number of moles of O2 lost using its molar mass: 0.302 g / 32 g/mol = 0.0094375 mol.
- From the balanced equation, we know that 2 moles of Mg(ClO3)2 produce 3 moles of O2, so we can calculate the number of moles of Mg(ClO3)2: (2/3) * 0.0094375 mol = 0.00629167 mol.
2. Determine the total mass of the mixture:
- We know that the mass of the mixture is 1.890 grams.
3. Calculate the percentage of Mg(ClO3)2 in the mixture:
- Divide the initial mass of Mg(ClO3)2 by the total mass of the mixture and multiply by 100: (0.00629167 g / 1.890 g) * 100 = 0.333%.
Therefore, the percentage of Mg(ClO3)2 in the mixture is approximately 0.333%.