Answer:
when 445 grams of H2O are reacted, 4,300 grams (or 4.3 kilograms) of MnO2 are produced.
Step-by-step explanation:
The balanced chemical equation for the reaction is:
16H+ (aq) + 2MnO4- (aq) + 10Br- (aq) → 2MnO2 (s) + 5Br2 (aq) + 8H2O (l)
According to the equation, 2 moles of MnO4- react to produce 2 moles of MnO2. Therefore, we need to find out the number of moles of MnO4- that react with the given amount of H2O and then use stoichiometry to calculate the number of moles of MnO2 produced.
First, let's calculate the number of moles of H2O:
mass of H2O = 445 g
molar mass of H2O = 18.015 g/mol
number of moles of H2O = mass/molar mass = 445 g/18.015 g/mol = 24.7 mol
From the balanced equation, we can see that 1 mole of H2O reacts with 2 moles of MnO4-. Therefore, the number of moles of MnO4- required to react with 24.7 mol of H2O is:
number of moles of MnO4- = 2 × number of moles of H2O = 49.4 mol
Since 2 moles of MnO4- produce 2 moles of MnO2, we can say that 1 mole of MnO4- produces 1 mole of MnO2. Therefore, the number of moles of MnO2 produced is also 49.4 mol.
Finally, we can calculate the mass of MnO2 produced:
mass of MnO2 = number of moles of MnO2 × molar mass of MnO2
mass of MnO2 = 49.4 mol × 86.94 g/mol = 4,300 g
Therefore, when 445 grams of H2O are reacted, 4,300 grams (or 4.3 kilograms) of MnO2 are produced.