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Alborosie
3 years ago
11

A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen. After cooling to room temperature, thr

residual gas was passed through aqueous potassium hydroxide. What volume of gas was absorbed by the alkali?. (need the and with working quick)
Chemistry
1 answer:
Tpy6a [65]3 years ago
5 0
In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
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An alkaline battery produces electrical energy according to the following equation.
Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

          1 mol     2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

       Zn(s)          +      2 MnO₂(s)    +    H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol)    2 mol (87 g/mol)

       65 g                        174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

           65 g Zn react ≅ 174 g of MnO₂

            X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

           X g Zn react = 65 g x 21.5 g / 174 g

           X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂  produced = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

        no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

        no. of moles = 21.5 g / 87 g/mol

        no. of moles = 0.25 moles

Now,

Look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

                        2 mol                                       1 mol  

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

         2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

            0.25 moles of MnO₂  ≅ X mole of Zn(OH)₂

Do cross multiplication

           X mole of Zn(OH)₂ = 1 mole x 0.25 mol x  / 2 mol

         X mole of Zn(OH)₂  ≅ 0.125

Now Conver moles of  Zn(OH)₂  to mass

Formula used

         mass in grams = no. of moles x molar mass

Molar mass of  Zn(OH)₂

Molar mass of  Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of  Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of  Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

        mass in grams = 0.125 mol x 99.4 g/mol  

        mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂  will be produce.

5 0
3 years ago
A person lying out in the sun to get a tan is an example of conduction, convection, or radiation.
steposvetlana [31]

Answer:

its radiation

Explanation:

just googled it rn trust me

3 0
3 years ago
Read 2 more answers
What geologic features might form in convergent boundaries<br>Pls help
lisov135 [29]
1) deep trenches
2) active volcanoes
3) arcs of islands along the tectonic boundaries
4) in some cases, mountain ranges
5 0
3 years ago
How does the electric force between two charged particles change if the
Anuta_ua [19.1K]
C. It is decreased by a factor of 3.
3 0
3 years ago
Part C: complete the third column <br> Part D: complete the fourth column
Helga [31]

Answer:

Part C: P2 = 0.30 atm

Part D: V1 = 16.22 L.

Explanation:

Part C:

Initial pressure (P1) = 2.67 atm

Initial volume (V1) = 5.54 mL

Final pressure (P2) =.?

Final volume (V2) = 49 mL

The final pressure (P2) can be obtained as follow:

P1V1 = P2V2

2.67 x 5.54 = P2 x 49

Divide both side by 49

P2 = (2.67 x 5.54)/49

P2 = 0.30 atm

Therefore, the final pressure (P2) is 0.30 atm

Part D:

Initial pressure (P1) = 348 Torr

Initial volume (V1) =?

Final pressure (P2) = 684 Torr

Final volume (V2) = 8.25 L

The initial volume (V1) can be obtained as follow:

P1V1 = P2V2

348 x V1 = 684 x 8.25

Divide both side by 348

V1 = (684 x 8.25)/348

V1 = 16.22 L

Therefore, the initial volume (V1) is 16.22 L

6 0
3 years ago
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