Chlorine gas can be prepared according to the reaction: 16 HCl+2KMnO4->2MnCl2+8H2O How many grams of excess reactant in 25gHCI reacted with 25g KMnO4 are left unreacted
To find how much of the excess reactant is left unreacted, we first need to determine the limiting reactant.
Assuming that we have an excess of HCl and 25 g of each reactant:
- The molar mass of HCl is 36.5 g/mol.
- The molar mass of KMnO4 is 158.034 g/mol.
Now we can calculate the number of moles of each reactant:
- Moles of HCl = 25 g / 36.5 g/mol = 0.685 mol
- Moles of KMnO4 = 25 g / 158.034 g/mol = 0.158 mol
Looking at the balanced equation, we can see that the ratio of HCl to KMnO4 is 16:2, or 8:1. This means that for every 8 moles of HCl, we need 1 mole of KMnO4 for complete reaction.
To determine the limiting reactant, we will compare the moles of each reactant. Since 0.685 mol of HCl is greater than 0.158 mol of KMnO4, HCl is in excess.
To find out how much of the excess reactant is left unreacted, we need to calculate how many moles of HCl reacted. From the balanced equation, we know that 16 moles of HCl react with 2 moles of KMnO4.
Since we have 0.158 mol of KMnO4, we can set up a ratio:
0.158 mol KMnO4 / 2 mol KMnO4 = 0.079 mol HCl will react
Therefore, the excess amount of HCl is:
0.685 mol - 0.079 mol = 0.606 mol
To convert this to grams, we can use the molar mass of HCl:
0.606 mol HCl * 36.5 g/mol = 22.113 g
Therefore, 22.113 grams of the excess HCl will be left unreacted.