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.
A) Yes, 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, we can use the concept of stoichiometry. Stoichiometry relates the amounts of reactants and products in a chemical reaction.
1. Start by calculating the moles of O2 lost. To do this, we need to know the molar mass of O2, which is approximately 32 g/mol.
Given:
Mass of O2 lost = 0.302 g
Molar mass of O2 = 32 g/mol
Moles of O2 lost = (0.302 g) / (32 g/mol) = 0.0094375 mol
2. Next, we need to determine the stoichiometric ratio between O2 and Mg(ClO3)2. From the balanced equation, we know that for every 3 moles of O2, 2 moles of Mg(ClO3)2 are required.
Stoichiometric ratio = 2 moles Mg(ClO3)2 / 3 moles O2
3. Now, we can calculate the moles of Mg(ClO3)2 present.
Moles of Mg(ClO3)2 = (Stoichiometric ratio) × (Moles of O2 lost)
= (2/3) × (0.0094375 mol)
= 0.006291 mol
4. Finally, we can find the percentage of Mg(ClO3)2 in the mixture.
Percentage of Mg(ClO3)2 = (Moles of Mg(ClO3)2 / Total moles in the mixture) × 100
The total moles in the mixture can be calculated by dividing the mass of the mixture by its molar mass.
Given:
Mass of the mixture = 1.890 g
Molar mass of Mg(ClO3)2 = 183.2 g/mol
Total moles in the mixture = (Mass of the mixture) / (Molar mass of Mg(ClO3)2)
= (1.890 g) / (183.2 g/mol)
= 0.0103098 mol
Percentage of Mg(ClO3)2 = (0.006291 mol / 0.0103098 mol) × 100
= 61.01%
Therefore, the percentage of Mg(ClO3)2 in the mixture is approximately 61.01%.