Answer:
if 49.4 moles of MnO4- are available, 4300 g (or 4.3 kg) of MnO2 will be produced.
Explanation:
16H+ (aq) + 2MnO4- (aq) + 10Br- (aq) → 2MnO2 (s) + 5Br2 (aq) + 8H2O (l)
From the balanced equation, we can see that 2 moles of MnO4- reacts with 2 moles of MnO2. Therefore, the number of moles of MnO2 formed is equal to the number of moles of MnO4- used.
First, let's calculate the number of moles of MnO4- used:
m(H2O) = 445 g
M(H2O) = 18.015 g/mol
n(H2O) = m/M = 445 g / 18.015 g/mol = 24.7 mol
Since 1 mol of H2O reacts with 2 moles of MnO4-, the number of moles of MnO4- required to react with 24.7 mol of H2O is:
n(MnO4-) = 2 × n(H2O) = 49.4 mol
Therefore, if we have 49.4 moles of MnO4- available, all of the MnO4- will be consumed and form the same number of moles of MnO2. The molar mass of MnO2 is 86.94 g/mol. The mass of MnO2 produced can be calculated as:
m(MnO2) = n(MnO2) × M(MnO2) = 49.4 mol × 86.94 g/mol = 4300 g
Therefore, if 49.4 moles of MnO4- are available, 4300 g (or 4.3 kg) of MnO2 will be produced.