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wolverine [178]
3 years ago
5

Strontium and chlorine combine in only one ratio: one strontium atom for every two chlorine atoms. Based on this

Chemistry
1 answer:
Vinvika [58]3 years ago
4 0

The combinations of Strontium and chlorine  that are possible are only those in which strontium and chlorine combine in the ratio of 1:2.

1) The possible combinations are

  • 4 strontium atoms and 8 chlorine
  • 2.7 * 10^19 strontium atoms and5.4 * 10^19 chlorine atoms

2) The combinations that are not possible are

  • 20 strontium atoms and 60 chlorine atoms
  • 130 billion strontium atoms and 195 billion chlorine

We have to work out the ratio of Strontium and chlorine in each of the given  combinations in the question. Only the combinations in which the ratio of Strontium and chlorine is 1:2 is possible.

First case:

4 strontium atoms and 8 chlorine atoms gives a Strontium and chlorine ratio of 1:2 so it is possible.

Second case:

20 strontium atoms and 60 chlorine atoms gives a Strontium and chlorine ratio of 1:3 hence it is not possible.

Third case:

2.7 * 10^19 strontium atoms and 5.4 * 10^19 chlorine atoms gives a Strontium and chlorine ratio of 1:2 hence it is possible.

Fourth case:

130 billion strontium atoms and 195 billion chlorine atoms gives a Strontium and chlorine ratio of 1:1.5 hence it is not possible.

Learn more: brainly.com/question/9743981

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane is
DanielleElmas [232]

Answer:

6g

Explanation:

Step 1:

The balanced equation for the reaction between gaseous ethane and gaseous oxygen. This is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Step 2:

Determination of the masses of C2H6 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol

Mass of C2H6 from the balanced equation = 2 x 30 = 60g

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Mass of O2 from the balanced equation = 7 x 32 = 224g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 4 x 44 = 176g

From the balanced equation above,

60g of C2H6 reacted with 224g of O2 to produce 176g of CO2.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting reactant as it will be needed to obtain the maximum mass of CO2.

From the balanced equation above,

60g of C2H6 reacted with 224g of O2.

Therefore, 2.71g of C2H6 will react with = (2.71 x 224)/60 = 10.12g of O2.

From the above calculation, we can see that a higher mass of O2 is needed to react with 2.71g of C2H6, therefore, O2 is the limiting reactant.

Step 4:

Determination of the maximum mass of CO2 produced when 2.71 g of ethane is mixed with 7.6 g of oxygen.

The limiting reactant is used to determine the maximum mass.

From the balanced equation above,

224g of O2 produce 176g of CO2.

Therefore, 7.6g of O2 will produce = (7.6 x 176)/224 = 5.97g ≈ 6g of CO2

From the calculations made above, the maximum mass of CO2 produced is 5.97 ≈ 6g

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B is your answer to this
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3 years ago
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From  the  equation  above   the  reacting   ratio  of  KClO3   to  O2  is  2:3 therefore  the  number  of  moles  of  oxygen  produced  is  ( 4 x3)/2 =  6 moles  since   four  moles  of  KClO3  was  consumed
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