Mg(ClO32=MgCl2+3O2
12.omolMg(ClO3)2produces molMgCl2and MolO2
3.00 molofMg(ClO3)2produces what mass of O2
Same as your other stoichiometry problems.
To determine the mass of O2 produced when 3.00 mol of Mg(ClO3)2 is reacted, we need to use the stoichiometry of the balanced chemical equation.
Given:
Mg(ClO3)2 = MgCl2 + 3O2
Molar mass of O2 = 32.00 g/mol
Step 1: Calculate the molar mass of Mg(ClO3)2
Molar mass of Mg = 24.31 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Mg(ClO3)2 = (24.31 g/mol + 3*(35.45 g/mol) + 6*(16.00 g/mol)) = 223.22 g/mol
Step 2: Use the stoichiometry of the balanced chemical equation to determine the number of moles of O2 produced.
According to the balanced equation:
1 mol of Mg(ClO3)2 produces 3 mol of O2.
Since we have 3.00 mol of Mg(ClO3)2, the number of moles of O2 produced is:
3.00 mol Mg(ClO3)2 * (3 mol O2/1 mol Mg(ClO3)2) = 9.00 mol O2
Step 3: Calculate the mass of O2 produced.
Mass = Number of moles * Molar mass
Mass of O2 = 9.00 mol * 32.00 g/mol = 288.00 g
Therefore, when 3.00 mol of Mg(ClO3)2 is reacted, it produces 9.00 mol of O2, which is equivalent to 288.00 g of O2.