Chlorine gas can be prepared according to the reaction: 16HCI+2KMnO4->5CI2+2MnCI2+8H2O (a) how many moles of MnCI2 can be produced when 25g KMnO4 are mixed with 85g HCI?
To find the number of moles of MnCI2 produced, we need to calculate the moles of KMnO4 and HCI first.
1. Calculate the moles of KMnO4:
Given:
Mass of KMnO4 = 25g
Molar mass of KMnO4 = 39.1g/mol + 54g/mol + 16g/mol × 4 = 158.9g/mol
Moles of KMnO4 = Mass of KMnO4 / Molar mass of KMnO4
= 25g / 158.9g/mol
≈ 0.1575 mol
2. Calculate the moles of HCI:
Given:
Mass of HCI = 85g
Molar mass of HCI = 1g/mol + 35.5g/mol = 36.5g/mol
Moles of HCI = Mass of HCI / Molar mass of HCI
= 85g / 36.5g/mol
≈ 2.3288 mol
Now, we can use the stoichiometric coefficients from the balanced equation to determine the ratio of moles of MnCI2 produced to moles of KMnO4 consumed.
From the balanced equation: 2KMnO4 : 2MnCI2
Therefore, the ratio is 1:1, meaning that for every 1 mole of KMnO4 consumed, 1 mole of MnCI2 is produced.
Since we have determined that the moles of KMnO4 is approximately 0.1575 mol, we can conclude that the number of moles of MnCI2 produced is also 0.1575 mol.