2KMnO4(aq)+16HCl(aq)->2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)

A.)the volume of 0.414 M KMnO4 needed to produce 1.66 mol MnCl2

B.) the volume of 2.15 M HCl needed to produce 112 mL of 0.618 M KCl

C.)the molarity (M) of the HCl solution when 26.42 mL react with 28.38 mL of 0.398 M KMnO4

D.)the liters of Cl2 gas at STP produced by the reaction of 132 mL of 2.7 M HCl

E.)the liters of Cl2 gas at STP produced by the reaction of 24.0 mL of 0.691 M HCl and 15.6 mL of 0.566 M KMnO4

A.) To find the volume of 0.414 M KMnO4 needed to produce 1.66 mol MnCl2, we first need to determine the moles of KMnO4 needed for the reaction. From the balanced equation, we can see that 1 mole of KMnO4 produces 1 mole of MnCl2:

1.66 mol MnCl2 * (1 mol KMnO4 / 1 mol MnCl2) = 1.66 mol KMnO4

Now we can use the molarity of the KMnO4 solution to find the volume needed:

1.66 mol KMnO4 / 0.414 M = 4.01 L

So, 4.01 L of 0.414 M KMnO4 is needed to produce 1.66 mol MnCl2.

B.) From the balanced equation, 16 moles of HCl are needed to produce 2 moles of KCl. First, we need to determine the moles of KCl in the 112 mL of 0.618 M KCl solution:

112 mL * (1 L / 1000 mL) * 0.618 mol/L = 0.0692 mol KCl

Now we can find the moles of HCl needed:

0.0692 mol KCl * (16 mol HCl / 2 mol KCl) = 0.5536 mol HCl

Finally, we can find the volume of 2.15 M HCl needed:

0.5536 mol HCl / 2.15 M = 0.2574 L = 257.4 mL

Therefore, the volume of 2.15 M HCl needed to produce 112 mL of 0.618 M KCl is 257.4 mL.

C.) From the balanced equation, 16 moles of HCl react with 2 moles of KMnO4. First, we need to find the moles of KMnO4 in 28.38 mL of 0.398 M KMnO4 solution:

28.38 mL * (1 L / 1000 mL) * 0.398 mol/L = 0.01131 mol KMnO4

Now we can find the moles of HCl needed:

0.01131 mol KMnO4 * (16 mol HCl / 2 mol KMnO4) = 0.09048 mol HCl

Finally, we can find the molarity of the HCl solution:

0.09048 mol HCl / (26.42 mL * (1 L / 1000 mL)) = 3.42 M

Therefore, the molarity of the HCl solution is 3.42 M.

D.) From the balanced equation, 16 moles of HCl produce 5 moles of Cl2. First, we need to find the moles of HCl in 132 mL of 2.7 M HCl solution:

132 mL * (1 L / 1000 mL) * 2.7 mol/L = 0.3564 mol HCl

Now we can find the moles of Cl2 produced:

0.3564 mol HCl * (5 mol Cl2 / 16 mol HCl) = 0.1115 mol Cl2

At STP (0°C and 1 atm), 1 mole of any gas occupies 22.4 liters. So, the volume of Cl2 gas produced at STP is:

0.1115 mol Cl2 * 22.4 L/mol = 2.498 L

Therefore, 2.498 liters of Cl2 gas are produced at STP when 132 mL of 2.7 M HCl reacts.

E.) First, we need to determine the limiting reactant. We can find the moles of both reactants:

Moles of HCl: 24.0 mL * (1 L / 1000 mL) * 0.691 mol/L = 0.01658 mol HCl
Moles of KMnO4: 15.6 mL * (1 L / 1000 mL) * 0.566 mol/L = 0.00883 mol KMnO4

Now we convert these moles to moles of Cl2:

Moles of Cl2 from HCl: 0.01658 mol HCl * (5 mol Cl2 / 16 mol HCl) = 0.00518 mol Cl2
Moles of Cl2 from KMnO4: 0.00883 mol KMnO4 * (5 mol Cl2 / 2 mol KMnO4) = 0.022075 mol Cl2

Since 0.00518 mol Cl2 is less than 0.022075 mol Cl2, HCl is the limiting reactant. So, the moles of Cl2 produced are 0.00518 mol.

Now we can find the volume of Cl2 gas produced at STP:

0.00518 mol Cl2 * 22.4 L/mol = 0.116 L

Therefore, 0.116 liters of Cl2 gas are produced at STP when 24.0 mL of 0.691 M HCl and 15.6 mL of 0.566 M KMnO4 reacts.