Chlorine gas can be prepared according to the reaction: 16 HCI+ 2KMnO4-> 5CI2+KCI+2KCI2+8H2O (a) How many moles of MnCI can be produced when 25kg KMnO4 are mixed with 85g HCI?
To determine the number of moles of MnCI that can be produced, we need to calculate the limiting reactant. This is the reactant that will be completely consumed and therefore determines the maximum amount of product that can be formed.
First, let's convert the given mass of KMnO4 from kilograms to grams:
25 kg KMnO4 = 25000 g KMnO4
Next, let's calculate the number of moles of KMnO4:
Molar mass of KMnO4 = 39.1 g/mol (K) + 54.9 g/mol (Mn) + 64.0 g/mol (O) x 4 = 158.8 g/mol
moles of KMnO4 = mass of KMnO4 / molar mass of KMnO4
moles of KMnO4 = 25000 g KMnO4 / 158.8 g/mol = 157.5 mol KMnO4
Now, let's convert the given mass of HCI from grams to moles:
mass of HCI = 85 g HCI
Molar mass of HCI = 1.0 g/mol (H) + 35.5 g/mol (Cl) = 36.5 g/mol
moles of HCI = mass of HCI / molar mass of HCI
moles of HCI = 85 g HCI / 36.5 g/mol = 2.33 mol HCI
Next, we need to determine the stoichiometric ratio between KMnO4 and HCI. From the balanced equation, we can see that the ratio is 16:2, which simplifies to 8:1. This means that for every 8 moles of HCI, 1 mole of KMnO4 is required.
So, let's calculate the moles of KMnO4 needed for the reaction:
moles of KMnO4 needed = moles of HCI / 8
moles of KMnO4 needed = 2.33 mol HCI / 8 = 0.292 mol KMnO4
Since this is less than the moles of KMnO4 available (0.292 mol < 157.5 mol), the limiting reactant is HCI.
Finally, we can calculate the moles of MnCI that can be produced. From the balanced equation, we can see that the stoichiometric ratio between KMnO4 and MnCI is 2:1. This means that for every 2 moles of KMnO4, 1 mole of MnCI is produced.
moles of MnCI = moles of KMnO4 needed x (1 mol MnCI / 2 mol KMnO4)
moles of MNCI = 0.292 mol KMnO4 x (1 mol MnCI / 2 mol KMnO4) = 0.146 mol MnCI
Therefore, 0.146 moles of MnCI can be produced when 25 kg of KMnO4 are mixed with 85 g of HCI.