1.) What mass in grams of solute is needed to prepare 325 mL of 5.30×102 M KMnO4

Question

1.) What mass in grams of solute is needed to prepare 325 mL of 5.30×102 M KMnO4? Express your answer with the appropriate units.

2.) 26.7 mol HCl in 90.0 L of solution. Express your answer with the appropriate units.

3.) 6.40×102 mol Li2CO3 in 696 mL of solution. Express your answer with the appropriate units.

4.) Ethane burns in air to form carbon dioxide and water. The equation is 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

Part A In the above equation, how many liters of oxygen (O2) are required to burn 6.0  L of ethane (C2H6) gas? Assume that both gases are measured at the same temperature and pressure.Express your answer with the appropriate units.

Part B 2C2H6(g)+7O2(g)4CO2(g)+6H2O(g) In the equation above, what volume of ethane (C2H6) in liters is required to form 24  L of carbon dioxide (CO2)? Express your answer with the appropriate units.

5.) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) In the equation, if 14 L of ethane (C2H6) and 14 L of oxygen (O2) combined and burned to completion, which gas will be leftover after the reaction, and what is the volume of that gas remaining? Enter the name of the gas followed by its volume in liters to the nearest whole number, separated by a comma.

Explanation / Answer

1) We want to prepare 325 mL of 5.30 x 10–2 M KMnO4 solution.

Let, for the KMnO4 solution to be prepared

Volume = 325 mL = 0.325 L ………….( since 1 mL = 1/1000 L)

Molarity = 5.30 x 10–2 M

From this we can calculate number of moles of KMnO4.

Number of moles = Molarity x Volume in L

Number of moles of KMnO4 in solution to be prepared = 5.30 x 10–2 x 0.325 = 1.7225 x 10–2 moles of KMnO4.

We know that ,

1 mole of substance = molar mass of substance

Molar mass of KMnO4 = 158.034 g/mol.

Hence we write correspondence as,

If,        1 mole of KMnO4 158.034 g KMnO4

Then, 1.7225 x 10–2 moles of KMnO4. say ‘M’ g KMnO4.

By cross multiplication way,

M x 1 = 1.7225 x 10–2 x 158.034

M = 2.722

I.e. 2.722 g of KMnO4 need to be dissolved and diluted to 345 mL distilled water which then gives 5.30 x 10–2 M KMnO4 solution.

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2)

Number of moles = Molarity in M/L x Volume in L

For HCl given that,

Number of moles = 26.7

Volume = 90.0 L

Molarity in M/L = ?

Let us put these values in above equation and calculate unknown part,

26.7 = Molarity in M/L x 90.0

Molarity in M/L = 26.7/90

Molarity in M/L = 0.297 M/L

Concentration of HCl solution formed thus is 0.297 M/L.

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3)

Number of moles = Molarity in M/L x Volume in L

For Li2CO3 solution given that,

Number of moles = 6.40 x 10–2 mole.

Volume = 696 mL = 0.696 L …………….(1 mL = 1/1000 L)

Molarity in M/L =?

Let us put these values in above equation and calculate unknown part,

6.40 x 10–2 = Molarity in M/L x 0.696

Molarity in M/L = 6.40 x 10–2 /0.696

Molarity in M/L = 9.2 x 10–2 M/L.

Concentration of Li2CO3 solution formed thus is 9.2 x 10–2 M/L.

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4) Combustion reaction of Ethane is,

2C2H6 (g) + 7O2 (g) ---------> 4CO2 (g) + 6H2O (g)

Stoichiometry says

2 mol C2H6 (g) 7 mol O2 (g) 4 mol CO2 (g) 6H2O (g)

Avogadro’s hypothesis states that 1 mole of gas at STP conditions occupy 22.4 L of the volume.

Hence above correspondence in terms of volume is given as,

2 x 22.4 L C2H6 (g) 7 x 22.4 L O2 (g) 4 x 22.4 L CO2 (g) 6 x 22.4 L H2O (g)

44.8 L C2H6 (g) 156.8 L O2 (g) 89.6 L CO2 (g) 134.4 L H2O (g) ……………………(1)

Part-A)

Required correspondence is,

If,         44.8 L C2H6 (g) 156.8 L O2 (g)

Then, 6.0 L C2H6 (g) say V1 L O2 (g)

By cross multiplication method,

V1 x 44.8 = 156.8 x 6.0

V1 = 156.8 x 6.0 / 44.8

V1 = 21 L

For complete combustion of 6 L of Ethane it will require 21 L of O2 gas.

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Part-B

Required correspondence is,

If        44.8 L C2H6 (g) 89.6 L CO2 (g)

Then, say V2 L C2H6 (g) 24 L CO2 (g)

Hence,

V2 x 89.6 = 44.8 x 24

V2 = 24 x 44.8/89.6

V2 = 12 L

12 L of Ethane gas need to be burnt for 24 L Of CO2 formation.

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