Given that Mg(ClO3)2 decomposes into magnesium chloride and oxygen:
A) Write a balanced equation for this reaction.
I got Mg(ClO3)2 yields MgCl2 + 3 O2
Is that right?
B) 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?
Please help! I'm lost.
The equation is right.
Mg(ClO3)2 --> MgCl2 + 3O2
Convert g O2 to mols. mols = grams/molar mass = about(estimated)0.009 but you should do it more accurately.
Convert mols O2 to mols Mg(ClO3)2 using the coefficients in the balanced equation. That is
0.009 x (1 mol Mg(ClO3)2/3 mols O2) = 0.009 x (1/3) = about 0.003 (again an estimate).
Convert mols Mg(ClO3)2 to g. g = mols x molar mass
Then %Mg(ClO3)2 = (g Mg(ClO3)2/mass sample)*100 = ?
A) Your balanced equation for the decomposition of Mg(ClO3)2 is correct:
2 Mg(ClO3)2 → 2 MgCl2 + 3 O2
B) To find the percentage of Mg(ClO3)2 in the mixture, you need to determine the mass of Mg(ClO3)2 present in the given mixture.
1. Start by calculating the number of moles of O2 lost:
Given mass of O2 lost = 0.302 grams
Molar mass of O2 = 32 g/mol
Number of moles of O2 = mass / molar mass = 0.302 g / 32 g/mol = 0.00944 mol
2. Use the balanced equation to find the number of moles of Mg(ClO3)2 that would produce the same amount of O2:
From the balanced equation, 2 moles of Mg(ClO3)2 produce 3 moles of O2.
So, the number of moles of Mg(ClO3)2 = (0.00944 mol O2) × (2 mol Mg(ClO3)2 / 3 mol O2) = 0.00629 mol
3. Now, calculate the molar mass of Mg(ClO3)2:
Molar mass of Mg(ClO3)2 = (1 mol Mg × atomic mass of Mg) + (2 mol Cl × atomic mass of Cl) + (6 mol O × atomic mass of O)
= (1 mol × 24.31 g/mol) + (2 mol × 35.45 g/mol) + (6 mol × 16.00 g/mol)
= 24.31 g + 70.90 g + 96.00 g
= 191.21 g/mol
4. Finally, calculate the mass of Mg(ClO3)2 in the mixture:
Mass of Mg(ClO3)2 = number of moles × molar mass
= 0.00629 mol × 191.21 g/mol
= 1.204 g
5. Calculate the percentage of Mg(ClO3)2 in the mixture:
Percentage of Mg(ClO3)2 = (mass of Mg(ClO3)2 / mass of mixture) × 100%
= (1.204 g / 1.890 g) × 100%
= 63.75%
Therefore, the percentage of Mg(ClO3)2 in the mixture is approximately 63.75%.
A) Your balanced equation, Mg(ClO3)2 → MgCl2 + 3 O2, is correct. Well done!
B) To find the percentage of Mg(ClO3)2 in the mixture, you need to determine the mass of Mg(ClO3)2 and then divide it by the total mass of the mixture. Here's the step-by-step process:
Step 1: Calculate the mass of oxygen lost
Given: Mass of O2 = 0.302 grams
Since there are three moles of oxygen produced for every one mole of Mg(ClO3)2, we can use the molar mass of O2 to find the moles of oxygen produced:
Molar mass of O2 = 32 g/mol
Moles of O2 = Mass of O2 / Molar mass of O2
= 0.302 g / 32 g/mol
= 0.0094375 mol
Step 2: Calculate the moles of Mg(ClO3)2
To find the moles of Mg(ClO3)2, we can use stoichiometry from the balanced equation. From our equation, we know that one mole of Mg(ClO3)2 yields three moles of O2.
Moles of O2 = Moles of Mg(ClO3)2 / 3
0.0094375 mol = Moles of Mg(ClO3)2 / 3
Moles of Mg(ClO3)2 = 0.0094375 mol * 3
= 0.0283125 mol
Step 3: Calculate the mass of Mg(ClO3)2
Now, we can determine the mass of Mg(ClO3)2 using its molar mass:
Molar mass of Mg(ClO3)2 = (24.31 g/mol + 35.45 g/mol + (3 * (16 g/mol))) * 2
= 24.31 g/mol + 35.45 g/mol + 48 g/mol
= 107.76 g/mol
Mass of Mg(ClO3)2 = Moles of Mg(ClO3)2 * Molar mass of Mg(ClO3)2
= 0.0283125 mol * 107.76 g/mol
= 3.05 grams
Step 4: Calculate the percentage of Mg(ClO3)2
Now, to determine the percentage of Mg(ClO3)2, divide the mass of Mg(ClO3)2 by the total mass of the mixture and multiply by 100:
Percentage of Mg(ClO3)2 = (Mass of Mg(ClO3)2 / Total mass of the mixture) * 100
= (3.05 g / 1.890 g) * 100
= 161.375%
Therefore, the percentage of Mg(ClO3)2 in the mixture is approximately 161.375%.