Question

The reaction of hydrochloric acid with potassium permanganate is described above. 526.64 g of hydrogen chloride is reacted with 229.19 g of potassium permanganate. Assuming the reaction goes to completion, calculate the mass of each product produced. g of manganese(II) chloride g of water g of chlorine g of potassium chloride

Answer 1

Answer:

(1) mass of KCl =108.025g

(2) mass of MnCl2 =182.7 g

(3) mass of Cl2 =257.37 g

(4) mass of H20(water) =104.4g

Explanation:

we start with determining the limiting factor

molar mass of KMnO4 = 158g/mol

molar mass of HCl = 36.5g/mol

hence,

number of moles of KMnO4= mass /molar mass

number of moles of KMnO4= 229.19/158 = 1.45moles

number of moles of HCl = 526.64/36.5 = 14.43 moles

we chose the lowest number of moles from the reactants as the limiting factor

hence the limiting factor is KMnO4.

calculating the mass of the products:

for KCl

2moles of KMnO4 reacts to give 2moles of KCl

there for KCl contains 1.45moles

(1) mass of KCl = number of moles of KCl x molar mass of KCl = 1.45 x 74.5 = 108.025g

(2) from mole ratio 2 mole of KMnO4 gave 2 moles of MnCl2

therefore 1.45moles KMnO4 will give 1.45 MnCl2

mass of MnCl2 = 1.45moles x molar mass MnCl2 = 1.45 x 126 = 182.7 g

(3) from mole ratio

2moles KMnO4 gave 5moles Cl2

1.45 moles will give (5 x1.45)/2 moles of Cl2

mass of Cl2 = number of moles of Cl2 x molar mass Cl2 = 3.625 x 71 = 257.37 g

(4)from mole ratio

2moles KMnO4 gave 8moles water(H2O)

1.45 moles will give (1.45x 8)/2 moles of H20

mass of H2O produced = number of moles of H20 x the molar mass of H20 = 5.8 x 18 = 104.4g

Answer 2

Explanation:

Equation of the reaction:

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

To calculate the limiting reagent, we need to calculate the number of moles of the reactants :

KMnO4:

Molar mass = (39 + 55 + (16*4))

= 158 g/mol

Number of moles = mass/molar mass

= 229.19/158

= 1.4506 mol

HCl:

Molar mass = 1 + 35.5

= 36.5 g/mol

Number of moles = 526.64/36.5

= 14.428 mol

By stoichiometry, 2 moles of KMnO4 reacted with 16 moles of HCl

The limiting reagent :

14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl

= 1.8035 moles of KMnO4 is required to react with 14.428 moles of HCl

But there's 1.4506 moles of KMnO4. Therefore, KMnO4 is the limiting reagent.

Mass of the products:

KCl:

2 moles of KMnO4 will produce 2 moles of KCl

Moles of KCl = 1 * 1.4506 mol

= 1.4506 mol

Molar mass = 39 + 35.5 = 74.5 g/mol

Mass of KCl = 74.5 * 1.4506

= 108.07 g

MnCl2:

2 moles of KMnO4 will produce 2 moles of MnCl2

Number of moles of MnCl2 = 1 * 1.4506

= 1.4506 mol

Molar mass = 55 + (35.5*2)

= 126 g/mol

Mass of MnCl2= 1.4506 * 126

= 182.78 g

Cl2:

2 moles of KMnO4 will produce 5 moles of Cl2

Number of moles of Cl2 = 5/2 * 1.4506

= 3.6265 mol

Molar mass of Cl2 = 35.5*2

= 71 g/mol

Mass of Cl2 = 71* 3.6265

= 257.4815 g

H2O:

2 moles of KMnO4 will produce 8 moles of H2O

Number of moles of H2O = 8/2 * 1.4506

= 5.80 mol

Molar mass of H2O =( 1*2) + 16

= 18 g/mol

Mass of H2O = 18*5.80

= 104.44 g